Calculus — Derivatives

Introduction

The derivative of a function measures its instantaneous rate of change. Geometrically, $f'(a)$ is the slope of the tangent line to $y = f(x)$ at $x = a$ ; physically, it is the speed at which a quantity is changing at one instant rather than on average over an interval.

Derivatives power every optimisation problem, every physical law stated as a rate, and every approximation method in applied mathematics. The formal definition,

f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h},

captures a single idea: zoom in on a curve until it looks straight, then measure the slope of that line. Everything in this chapter follows from it. The first three sections say precisely what the definition means and when it fails; the middle of the chapter builds the rules that let you differentiate any function written down in closed form, proving each one; the last sections put the derivative to work.

2.1The derivative at a point and as a function

Definition 2.1 (Derivative at a point). Let $f$ be defined on an open interval containing $a$ . The derivative of $f$ at $a$ is

f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h},

whenever this limit exists, in which case $f$ is said to be differentiable at $a$ . Writing $x = a + h$ , the same limit reads

f'(a) = \lim_{x\to a} \frac{f(x)-f(a)}{x-a}.

The quotient inside the limit is the slope of the secant line through the two points $(a, f(a))$ and $(a+h, f(a+h))$ on the graph of $f$ : rise over run. As $h \to 0$ the second point slides toward the first, the secant pivots into the tangent, and its slope becomes the derivative. The two forms of the definition are the same limit in different clothes; the $h$ -form is usually better for algebra, the $x \to a$ form for theory.

Two hypotheses are doing quiet work. The function must be defined on both sides of $a$ , because $h$ approaches $0$ through positive and negative values, and the limit must be a single finite number. The next two sections are about what happens when either of these fails.

The definition produces a number at each point where it succeeds. Letting the point vary turns those numbers into a new function.

Definition 2.2 (The derivative as a function). The derivative of $f$ is the function

f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h},

whose domain is the set of $x$ at which the limit exists. That domain may be smaller than the domain of $f$ .

Intuition. A speedometer reports the derivative of your position. Average speed over a trip is distance divided by time — a secant slope. The speedometer needle is what that ratio tends to as the trip is shortened to an instant: $50$ km in $1$ hour, then $0.8$ km in $1$ minute, then $14$ metres in $1$ second, each ratio settling toward the same $50$ km/h. The limit in the definition is that settling.

Example 2.3 (A derivative from the definition). Find $f'(x)$ for $f(x) = x^{2} + 3x$ directly from the definition.

Solution.

Expand the numerator: $f(x+h) = (x+h)^2 + 3(x+h) = x^{2} + 2xh + h^{2} + 3x + 3h$ , so $f(x+h)-f(x) = 2xh + h^{2} + 3h$ .
Divide by $h$ , which is legitimate because $h \neq 0$ inside the limit:

\frac{2xh + h^2 + 3h}{h} = 2x + h + 3.

Let $h \to 0$ : $f'(x) = 2x + 3$ .
Sanity check: at $x = 0$ the slope should equal the coefficient of $x$ , which is $3$ , and $f'(0) = 3$ .

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Example 2.4 (A square root from the definition). Find $f'(x)$ for $f(x) = \sqrt{x}$ , and state the domain of $f'$ .

Solution.

The difference quotient is $\dfrac{\sqrt{x+h} - \sqrt{x}}{h}$ , of the form $\frac{0}{0}$ as $h \to 0$ .
Multiply by the conjugate:

\frac{\sqrt{x+h} - \sqrt{x}}{h}\cdot\frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} = \frac{(x+h) - x}{h\left(\sqrt{x+h}+\sqrt{x}\right)} = \frac{1}{\sqrt{x+h}+\sqrt{x}}.

Let $h \to 0$ : $f'(x) = \dfrac{1}{2\sqrt{x}}$ .
The domain of $f$ is $[0,\infty)$ , but the domain of $f'$ is $(0,\infty)$ : at $x = 0$ the quotient $\frac{1}{\sqrt{h}}$ blows up, and the graph has a vertical tangent there.

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Example 2.5 (The $x \to a$ form). Find the slope of the tangent to $y = \dfrac{1}{x}$ at the point $\left(2, \tfrac12\right)$ .

Solution.

Use the second form of the definition with $a = 2$ :

f'(2) = \lim_{x\to 2}\frac{\frac1x - \frac12}{x - 2}.

Combine the numerator over a common denominator: $\frac1x - \frac12 = \frac{2 - x}{2x}$ , so the quotient is

\frac{2-x}{2x(x-2)} = -\frac{1}{2x}.

Let $x \to 2$ : $f'(2) = -\frac14$ .
Sanity check: $1/x$ is decreasing, so the slope must be negative, and it is.

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Pitfall. In the difference quotient, $f(x+h)$ means "substitute $x+h$ everywhere $x$ appears". It does not mean $f(x) + h$ . For $f(x) = x^{2}$ the numerator is $(x+h)^2 - x^{2} = 2xh + h^{2}$ , not $h$ .

Exercises for §2.1 →

2.2Notation

Three notations for the derivative are in daily use, and each is best at something.

Notation (Lagrange, Leibniz and operator notation). For $y = f(x)$ the following all denote the derivative:

f'(x) = y' = \frac{dy}{dx} = \frac{d}{dx}f(x) = Df(x) = D_x f(x).

The value at a particular point $a$ is written $f'(a)$ or, in Leibniz notation,

\left.\frac{dy}{dx}\right|_{x=a}.

In physics, a derivative with respect to time is often written with a dot: $\dot{s} = \dfrac{ds}{dt}$ .

Lagrange's prime is the most compact and the natural choice for a function of one variable. Leibniz's $\frac{dy}{dx}$ is longer but carries more information: it names both the dependent and the independent variable, which matters as soon as several variables are in play, and it looks like a fraction because it comes from one — the difference quotient $\frac{\Delta y}{\Delta x}$ with $\Delta x \to 0$ . Several rules later in the chapter are easiest to remember in this form; the chain rule reads

\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx},

as if the $du$ 's cancelled.

Take that cancellation as a memory aid, not a fact. On its own $\frac{dy}{dx}$ is a single symbol for a limit; the pieces $dy$ and $dx$ are not yet numbers. They can be given a meaning of their own — as differentials, later in this chapter — in such a way that the fraction of the two differentials really does equal the derivative. Until then, treat $\frac{d}{dx}$ as an operator: a machine that takes a function and returns its derivative. Writing $\frac{d}{dx}\left[x^{3}\right] = 3x^{2}$ says exactly that.

The second derivative, the derivative of the derivative, is written $f''(x)$ , $y''$ or

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right),

the exponent $2$ recording that the operator $\frac{d}{dx}$ has been applied twice. The placement is deliberate: $d^{2}y$ on top, $dx^{2}$ below, because two $d$ 's act on $y$ while $(dx)^2$ is a single small quantity squared.

Example 2.6 (Reading a statement in each notation). The tangent to $y = x^{3}$ at $x = 2$ has slope $12$ . Write this in Lagrange, Leibniz and operator notation.

Solution.

Lagrange: with $f(x) = x^{3}$ , $f'(2) = 12$ .
Leibniz: $\left.\dfrac{dy}{dx}\right|_{x=2} = 12$ .
Operator: $\left.\dfrac{d}{dx}\left[x^{3}\right]\right|_{x=2} = 3x^{2}\big|_{x=2} = 12$ .
All three say the same thing; the operator form additionally shows the derivative as a function ( $3x^{2}$ ) before it is evaluated.

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Pitfall. $\frac{dy}{dx}$ is not $d$ times $y$ over $d$ times $x$ , and the $d$ 's do not cancel to give $\frac{y}{x}$ . Equally, $\frac{d^{2}y}{dx^{2}}$ is not the square of $\frac{dy}{dx}$ : for $y = x^{3}$ the first is $6x$ and the second is $9x^{4}$ .

Exercises for §2.2 →

2.3Differentiability and continuity

Theorem 2.7 (Differentiability implies continuity). If $f$ is differentiable at $a$ , then $f$ is continuous at $a$ .

Proof. For $x \neq a$ write

f(x) - f(a) = \frac{f(x)-f(a)}{x-a}\cdot(x-a).

As $x \to a$ the first factor tends to $f'(a)$ and the second to $0$ , so by the product law for limits $f(x) - f(a) \to f'(a)\cdot 0 = 0$ . That is, $\lim_{x\to a} f(x) = f(a)$ , which is continuity at $a$ .∎

Read the proof as a statement about size: near $a$ the change in $f$ is roughly $f'(a)$ times the change in $x$ , so as the change in $x$ shrinks to nothing, so does the change in $f$ . A function with a jump cannot have a derivative at the jump, because the numerator of the difference quotient refuses to go to zero while the denominator does.

The converse is false, and the failures come in a small number of recognisable shapes. To describe them precisely we need the two halves of the limit in the definition.

Definition 2.8 (One-sided derivatives). The left-hand and right-hand derivatives of $f$ at $a$ are

f'_-(a) = \lim_{h\to 0^-}\frac{f(a+h)-f(a)}{h}, \qquad f'_+(a) = \lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}.

By the two-sided criterion for limits, $f'(a)$ exists exactly when both one-sided derivatives exist and are equal. A function is differentiable on an open interval when it is differentiable at every point of the interval.

A continuous function fails to be differentiable at $a$ in one of three ways.

A corner: both one-sided derivatives exist but differ. The model is $|x|$ at $0$ , where the difference quotient is $-1$ for $h < 0$ and $+1$ for $h > 0$ . The graph has two distinct tangent directions meeting at a point.

A vertical tangent: the difference quotient tends to $+\infty$ or $-\infty$ from both sides. The model is $\sqrt[3]{x}$ at $0$ , where the quotient is $h^{1/3}/h = h^{-2/3} \to +\infty$ . The graph is smooth to the eye, but its tangent is vertical and a vertical line has no slope.

A cusp: the one-sided quotients tend to infinity with opposite signs. The model is $x^{2/3}$ at $0$ , where the quotient is $h^{2/3}/h = h^{-1/3}$ , which tends to $-\infty$ from the left and $+\infty$ from the right. The graph comes to a sharp point with a vertical tangent on each side.

A discontinuity of any kind is the fourth way, and the theorem above says it is unavoidable: no derivative can exist there.

Intuition. Zoom in on a curve with a magnifying glass. A differentiable curve straightens out: at high enough magnification it is indistinguishable from a line, and the slope of that line is the derivative. A corner never straightens — at every magnification it still shows two directions. A vertical tangent straightens, but into a vertical line, which has no slope. A jump never even connects.

Example 2.9 (A corner, from the one-sided derivatives). Show that $f(x) = |x|$ is continuous but not differentiable at $0$ .

Solution.

Continuity: $|x| \to 0 = f(0)$ as $x \to 0$ , since $\big||x| - 0\big| = |x|$ .
Right-hand derivative: for $h > 0$ , $|h| = h$ , so the quotient is $\frac{h - 0}{h} = 1$ and $f'_+(0) = 1$ .
Left-hand derivative: for $h < 0$ , $|h| = -h$ , so the quotient is $\frac{-h}{h} = -1$ and $f'_-(0) = -1$ .
The one-sided derivatives differ, so $f'(0)$ does not exist. Away from $0$ the function is differentiable, with $f'(x) = 1$ for $x > 0$ and $f'(x) = -1$ for $x < 0$ .

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Example 2.10 (A vertical tangent). Show that $f(x) = \sqrt[3]{x}$ is not differentiable at $0$ , although it is differentiable everywhere else.

Solution.

At $0$ the difference quotient is

\frac{\sqrt[3]{h} - 0}{h} = \frac{h^{1/3}}{h} = \frac{1}{h^{2/3}},

which tends to $+\infty$ as $h \to 0$ from either side because $h^{2/3} > 0$ is small. The limit is not a finite number, so $f'(0)$ does not exist. 2. For $x \neq 0$ the power rule of the next section gives $f'(x) = \frac13 x^{-2/3}$ , which is finite. 3. The graph passes through the origin vertically: this is a vertical tangent, not a corner. The picture is smooth; only the slope is unavailable.□

Example 2.11 (Making a piecewise function differentiable). Let

f(x) = \begin{cases} x^2, & x \le 1 \\ mx + b, & x > 1. \end{cases}

Find $m$ and $b$ so that $f$ is differentiable at $x = 1$ .

Solution.

Differentiability requires continuity, so the two pieces must agree at $x = 1$ : $1^2 = m\cdot 1 + b$ , that is $m + b = 1$ .
The left-hand derivative comes from $x^{2}$ : $f'_-(1) = 2\cdot 1 = 2$ .
The right-hand derivative comes from the line: $f'_+(1) = m$ . Differentiability requires $m = 2$ .
Then $b = 1 - m = -1$ , so the line is $y = 2x - 1$ , which is exactly the tangent line to $y = x^{2}$ at $(1,1)$ . The parabola flows into its own tangent with no corner.
Sanity check: with $m = 3$ , $b = -2$ the function is still continuous but has a corner, since the slopes $2$ and $3$ disagree.

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Pitfall. Continuity is necessary for differentiability, not sufficient. Checking that two pieces of a function meet is only half the job; the slopes must also match from both sides.

Exercises for §2.3 →

2.4The basic rules: constants, powers and sums

Computing every derivative from the definition would be unbearable. The rest of the chapter replaces the definition with a short list of rules, each proved once from the definition and then used forever.

Theorem 2.12 (Constant, constant-multiple, sum and difference rules). Let $f$ and $g$ be differentiable at $x$ and let $c$ be a constant. Then

\frac{d}{dx}\,c = 0, \qquad \frac{d}{dx}\left[cf(x)\right] = cf'(x), \qquad \frac{d}{dx}\left[f(x) \pm g(x)\right] = f'(x) \pm g'(x).

Proof. For the constant function, $f(x+h) - f(x) = c - c = 0$ , so every difference quotient is $0$ and so is the limit.

For the constant multiple, $\dfrac{cf(x+h) - cf(x)}{h} = c\cdot\dfrac{f(x+h)-f(x)}{h}$ , and the constant-multiple law for limits gives $cf'(x)$ .

For the sum,

\frac{[f(x+h) + g(x+h)] - [f(x) + g(x)]}{h} = \frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h},

and the sum law for limits gives $f'(x) + g'(x)$ . The difference rule is the sum rule applied to $f + (-1)g$ .∎

Together these say that differentiation is linear: it passes through sums and through constant factors. In practice that is permission to differentiate a long expression term by term, dealing with each piece on its own.

Theorem 2.13 (Power rule for positive integers). For every positive integer $n$ ,

\frac{d}{dx}x^n = nx^{n-1}.

Proof. By the binomial theorem,

(x+h)^n = x^n + nx^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots + h^n.

Subtract $x^n$ and divide by $h$ : every term that survives carries at least one factor of $h$ except the second,

\frac{(x+h)^n - x^n}{h} = nx^{n-1} + \binom{n}{2}x^{n-2}h + \cdots + h^{n-1}.

As $h \to 0$ every term after the first vanishes, leaving $nx^{n-1}$ .∎

The proof explains the shape of the rule: differentiating asks how $x^n$ responds to a small nudge $h$ , and to first order the response is $n$ copies of $x^{n-1}h$ , one for each factor in the product $x\cdot x\cdots x$ that the nudge could land on.

The rule holds for every real exponent, not just positive integers, and we use that freely from here on: $\frac{d}{dx}x^{-2} = -2x^{-3}$ and $\frac{d}{dx}\sqrt{x} = \frac{d}{dx}x^{1/2} = \frac12 x^{-1/2}$ are the same rule. The general case has to wait for more tools — negative integers fall to the quotient rule, and arbitrary real exponents to logarithmic differentiation — but nothing in the meantime depends on the order of proof.

Intuition. Think of $x^{3}$ as the volume of a cube of side $x$ . Grow the side by a sliver $h$ and the volume grows by three thin slabs of area $x^{2}$ and thickness $h$ (one per face-pair), plus corner pieces of negligible size $h^{2}$ and $h^{3}$ . The growth per unit of $h$ is $3x^{2}$ : the power rule is a statement about how many faces a cube has.

Example 2.14 (The power rule on a polynomial). Differentiate $3x^{4} - 5x^{2} + 7x - 2$ .

Solution.

Linearity lets us take the four terms separately and pull the coefficients out front.
$\frac{d}{dx}(3x^{4}) = 3\cdot 4x^{3} = 12x^{3}$ : bring the exponent down, then reduce it by one.
$\frac{d}{dx}(-5x^{2}) = -10x$ and $\frac{d}{dx}(7x) = 7$ , since $x = x^{1}$ gives $1\cdot x^{0} = 1$ .
$\frac{d}{dx}(-2) = 0$ by the constant rule.
Adding up, the derivative is $12x^{3} - 10x + 7$ .

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Example 2.15 (Roots and reciprocals as powers). Differentiate $g(x) = \dfrac{4}{x^{3}} + 6\sqrt{x} - \dfrac{1}{\sqrt[3]{x}}$ .

Solution.

Rewrite every term as a power: $g(x) = 4x^{-3} + 6x^{1/2} - x^{-1/3}$ .
Apply the power rule term by term:

g'(x) = 4(-3)x^{-4} + 6\cdot\tfrac12 x^{-1/2} - \left(-\tfrac13\right)x^{-4/3} = -12x^{-4} + 3x^{-1/2} + \tfrac13 x^{-4/3}.

In the original notation, $g'(x) = -\dfrac{12}{x^{4}} + \dfrac{3}{\sqrt{x}} + \dfrac{1}{3x\sqrt[3]{x}}$ .
Sanity check on signs: $4/x^{3}$ decreases for $x > 0$ , and its derivative $-12/x^{4}$ is negative there.

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Example 2.16 (Tangents parallel to a given line). Find the points on the curve $y = x^{3} - 3x$ where the tangent line is parallel to $y = 9x + 1$ , and give the tangent lines.

Solution.

Parallel lines have equal slopes, so we need $y' = 9$ .
$y' = 3x^{2} - 3$ , and $3x^{2} - 3 = 9$ gives $x^{2} = 4$ , so $x = \pm 2$ .
The points are $(2, 8 - 6) = (2, 2)$ and $(-2, -8 + 6) = (-2, -2)$ .
Through $(2,2)$ with slope $9$ : $y - 2 = 9(x - 2)$ , that is $y = 9x - 16$ . Through $(-2,-2)$ : $y = 9x + 16$ .
Sanity check: the curve is odd, so its tangent lines come in symmetric pairs, and these two are reflections of each other through the origin.

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Pitfall. The power rule needs a constant exponent. It does not apply to $2^x$ , $x^x$ or $e^x$ , whose exponents move. Those need the exponential rule and logarithmic differentiation, both later in this chapter.

Exercises for §2.4 →

2.5The exponential function

Let $a > 0$ and try the definition on $f(x) = a^x$ :

f'(x) = \lim_{h\to 0}\frac{a^{x+h} - a^x}{h} = \lim_{h\to 0}\frac{a^x\left(a^h - 1\right)}{h} = a^x\cdot\lim_{h\to 0}\frac{a^h - 1}{h}.

The factor $a^x$ comes out of the limit because it does not depend on $h$ . What remains is a limit that depends only on the base — call it $L(a)$ — and it is exactly $f'(0)$ , the slope of the graph of $a^x$ as it crosses the $y$ -axis. So every exponential function has a derivative proportional to itself:

\frac{d}{dx}a^x = L(a)\,a^x, \qquad L(a) = \lim_{h\to 0}\frac{a^h-1}{h}.

Numerically, $L(2) \approx 0.693$ and $L(3) \approx 1.099$ : for $h = 0.001$ the quotients are $\frac{2^{0.001}-1}{0.001} \approx 0.6934$ and $\frac{3^{0.001}-1}{0.001} \approx 1.0992$ . Somewhere between $2$ and $3$ there is a base whose constant is exactly $1$ , and that base is the most convenient one in all of calculus.

Definition 2.17 (The number $e$ ). $e$ is the base for which the slope of $y = a^x$ at $x = 0$ is $1$ :

\lim_{h\to 0}\frac{e^h - 1}{h} = 1.

Its value is $e = 2.71828\ldots$

Theorem 2.18 (Derivative of $e^x$ ).

\frac{d}{dx}e^x = e^x.

Proof. Put $a = e$ in the computation above: $\frac{d}{dx}e^x = L(e)\,e^x = 1\cdot e^x$ .∎

This is the reason $e$ appears everywhere. The exponential $e^x$ is the function that equals its own rate of change: its height at any point is also its slope there. That makes it the natural language for anything whose growth is proportional to its size — populations, bank balances, radioactive samples — and it is why the limit in the Limits chapter, $\lim_{x\to 0}\frac{e^x-1}{x} = 1$ , is really a statement about a derivative.

For a general base the constant $L(a)$ turns out to be $\ln a$ , so that $\frac{d}{dx}a^x = a^x\ln a$ . Proving that needs the chain rule, and the proof is given in that section; the formula is stated here so you can use it right away.

Intuition. An account paying $100\%$ interest, compounded continuously, grows at a rate equal to its current balance: with $\$1{,}000$ in it, the balance grows at $\$1{,}000$ per year at that instant; with $\$2{,}000$ , at $\$2{,}000$ per year. That account balance is $1000e^t$ . A base of $2$ instead of $e$ is the same account growing at only $69.3\%$ of its balance per year, which is what $L(2) \approx 0.693$ measures.

Example 2.19 (Differentiating with $e^x$ ). Find $f'(x)$ for $f(x) = 3e^x - 2x^{4} + 5$ , and find $f'(0)$ .

Solution.

Linearity splits the three terms: $\frac{d}{dx}(3e^x) = 3e^x$ , $\frac{d}{dx}(-2x^{4}) = -8x^{3}$ , $\frac{d}{dx}(5) = 0$ .
So $f'(x) = 3e^x - 8x^{3}$ .
At $x = 0$ : $f'(0) = 3e^{0} - 0 = 3$ .

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Example 2.20 (A tangent to $y = e^x$ through the origin). Which tangent line to $y = e^x$ passes through the origin?

Solution.

The tangent at $x = a$ has slope $e^a$ and passes through $(a, e^a)$ : $y = e^a + e^a(x - a)$ .
It passes through $(0,0)$ when $0 = e^a - ae^a = e^a(1 - a)$ , and since $e^a \neq 0$ this forces $a = 1$ .
The tangent line is $y = e + e(x - 1) = ex$ .
Sanity check: the line $y = ex$ touches the curve at $(1, e)$ and has slope $e$ , which is $e^{1}$ , the derivative there.

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Example 2.21 (Matching a slope). At which point on $y = e^x$ is the tangent line parallel to $y = 2x$ ?

Solution.

We need $e^x = 2$ , since the derivative of $e^x$ is $e^x$ and the required slope is $2$ .
So $x = \ln 2$ and $y = e^{\ln 2} = 2$ : the point is $(\ln 2, 2)$ .
The tangent line there is $y - 2 = 2(x - \ln 2)$ . Notice that on $y = e^x$ the height and the slope always coincide, so "slope $2$ " and "height $2$ " are the same condition.

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Pitfall. $\frac{d}{dx}e^x = e^x$ , not $xe^{x-1}$ . The power rule is for a variable base with a fixed exponent; $e^x$ is a fixed base with a variable exponent, and the two rules are not interchangeable.

Exercises for §2.5 →

2.6Product rule

Theorem 2.22 (Product rule). If $f$ and $g$ are differentiable at $x$ , then so is $fg$ , and

(fg)'(x) = f'(x)g(x) + f(x)g'(x).

Proof. Subtract and add $f(x+h)g(x)$ in the numerator of the difference quotient:

\frac{f(x+h)g(x+h) - f(x)g(x)}{h} = \frac{f(x+h)g(x+h) - f(x+h)g(x)}{h} + \frac{f(x+h)g(x) - f(x)g(x)}{h}.

The first fraction is $f(x+h)\cdot\dfrac{g(x+h)-g(x)}{h}$ and the second is $g(x)\cdot\dfrac{f(x+h)-f(x)}{h}$ . As $h \to 0$ the difference quotients tend to $g'(x)$ and $f'(x)$ ; $g(x)$ is a constant; and $f(x+h) \to f(x)$ because $f$ , being differentiable, is continuous at $x$ . The limit is $f(x)g'(x) + g(x)f'(x)$ .∎

The proof is worth remembering for its one trick, adding and subtracting a mixed term, and for the moment at which it uses continuity: without differentiability implying continuity, the factor $f(x+h)$ would not be known to settle down. The derivative of a product is not the product of the derivatives; the symmetric two-term shape is a good self-check, since an answer that is not symmetric in the roles of $f$ and $g$ has dropped a term.

Before reaching for the rule, look for a simplification. The product $x^{3}\cdot x^{-1}$ is just $x^{2}$ , and differentiating that directly is faster and safer.

Intuition. A rectangle with sides $f$ and $g$ has area $fg$ . Grow both sides a little: the new area adds a strip of area $g\cdot\Delta f$ along one edge, a strip of area $f\cdot\Delta g$ along the other, and a tiny corner square of area $\Delta f\,\Delta g$ . Divide by the time the growth took and let it shrink: the two strips give $g f' + f g'$ , and the corner, being a product of two small quantities, vanishes. The product rule is the two strips.

Example 2.23 (Product rule with an exponential). Find $\frac{d}{dx}\left[x^{3} e^x\right]$ .

Solution.

Name the factors: $f(x) = x^{3}$ and $g(x) = e^x$ , with $f'(x) = 3x^{2}$ and $g'(x) = e^x$ .
Apply the rule: $(fg)' = f'g + fg' = 3x^{2}e^x + x^{3}e^x$ .
Factor the common $x^{2}e^x$ : $\frac{d}{dx}\left[x^{3}e^x\right] = x^{2}e^x(x+3)$ .
Sanity check at $x = 0$ : the formula gives $0$ , and directly, $x^{3}e^x$ is $x^{3}$ times something near $1$ near $0$ , whose slope at $0$ is $0$ .

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Example 2.24 (Checking the rule against expansion). Differentiate $y = (x^{2} + 1)(x^{3} - 2x)$ two ways.

Solution.

Product rule: $y' = 2x(x^{3} - 2x) + (x^{2}+1)(3x^{2} - 2) = 2x^{4} - 4x^{2} + 3x^{4} - 2x^{2} + 3x^{2} - 2 = 5x^{4} - 3x^{2} - 2$ .
Expand first: $y = x^{5} - 2x^{3} + x^{3} - 2x = x^{5} - x^{3} - 2x$ , so $y' = 5x^{4} - 3x^{2} - 2$ .
The two agree. When both factors are polynomials, expanding is often no harder; the product rule earns its keep when the factors are of different kinds, like $x^{3}$ and $e^x$ .

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Example 2.25 (Using values from a table). Suppose $f(2) = 3$ , $f'(2) = -1$ , $g(2) = 5$ and $g'(2) = 2$ . Find $(fg)'(2)$ and $\left(x^{2} f(x)\right)'$ at $x = 2$ .

Solution.

$(fg)'(2) = f'(2)g(2) + f(2)g'(2) = (-1)(5) + (3)(2) = 1$ .
For $x^{2} f(x)$ , the factors are $x^{2}$ and $f(x)$ : the derivative is $2x f(x) + x^{2} f'(x)$ .
At $x = 2$ : $2\cdot 2\cdot 3 + 4\cdot(-1) = 12 - 4 = 8$ .
No formula for $f$ was needed; the rule only consumes values and slopes at the point in question.

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Pitfall. $(fg)' \neq f'g'$ . Test it on $f(x) = g(x) = x$ : the product is $x^{2}$ with derivative $2x$ , while $f'g' = 1$ .

Exercises for §2.6 →

2.7Quotient rule

Theorem 2.26 (Quotient rule). If $f$ and $g$ are differentiable at $x$ and $g(x) \neq 0$ , then $f/g$ is differentiable at $x$ and

\left(\frac{f}{g}\right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}.

Proof. The difference quotient, after putting the two fractions over the common denominator $g(x+h)g(x)$ , is

\frac{1}{h}\left[\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}\right] = \frac{f(x+h)g(x) - f(x)g(x+h)}{h\,g(x+h)g(x)}.

Subtract and add $f(x)g(x)$ in the numerator, as in the product rule:

\frac{\left[f(x+h) - f(x)\right]g(x) - f(x)\left[g(x+h) - g(x)\right]}{h\,g(x+h)g(x)} = \frac{\dfrac{f(x+h)-f(x)}{h}\,g(x) - f(x)\,\dfrac{g(x+h)-g(x)}{h}}{g(x+h)g(x)}.

As $h \to 0$ the difference quotients tend to $f'(x)$ and $g'(x)$ , and $g(x+h) \to g(x)$ by continuity, so the limit is $\dfrac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ .∎

Order matters here in a way it does not for the product rule: $f'g$ comes first and $fg'$ is subtracted. The classroom mnemonic is "low d-high minus high d-low, over the square of what's below", where high is the numerator, low the denominator and d- means "derivative of". A quick check of the sign: $\frac{1}{x}$ is decreasing, and the rule with $f = 1$ , $g = x$ gives $\frac{0\cdot x - 1\cdot 1}{x^{2}} = -\frac{1}{x^{2}} < 0$ .

That last calculation generalises, and it pays off the first instalment on the general power rule.

Corollary 2.27 (Power rule for negative integers). For every positive integer $n$ and $x \neq 0$ , $\dfrac{d}{dx}x^{-n} = -nx^{-n-1}$ .

Proof. Apply the quotient rule to $x^{-n} = \dfrac{1}{x^n}$ :

\frac{d}{dx}\frac{1}{x^n} = \frac{0\cdot x^n - 1\cdot nx^{n-1}}{x^{2n}} = -nx^{n-1-2n} = -nx^{-n-1}.

∎

The rule earns its keep when the denominator is a genuine function, such as $\cos x$ or $e^x + 1$ . When the denominator is a single power, rewriting $f/g$ as $f\cdot g^{-1}$ , or just simplifying the algebra first, is usually cleaner.

Example 2.28 (Quotient rule with a trigonometric denominator). Find $\frac{d}{dx}\dfrac{x^{2}}{\cos x}$ .

Solution.

Identify the pieces: $f = x^{2}$ , $g = \cos x$ , $f' = 2x$ , $g' = -\sin x$ (proved in the next section).
Substitute into the rule:

\frac{f'g - fg'}{g^2} = \frac{2x\cos x - x^2(-\sin x)}{\cos^2 x} = \frac{2x\cos x + x^2\sin x}{\cos^2 x}.

Factor if you like: $\dfrac{x(2\cos x + x\sin x)}{\cos^{2} x}$ .

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Example 2.29 (A rational function). Differentiate $y = \dfrac{x^{2} + 1}{x^{2} - 1}$ and find the slope at $x = 0$ .

Solution.

$f = x^{2} + 1$ , $g = x^{2} - 1$ , $f' = g' = 2x$ .
Numerator of the rule: $2x(x^{2} - 1) - (x^{2}+1)(2x) = 2x\left[(x^{2} - 1) - (x^{2} + 1)\right] = -4x$ .
So $y' = \dfrac{-4x}{(x^{2}-1)^2}$ , and $y'(0) = 0$ .
Sanity check: the function is even, so its graph is symmetric about the $y$ -axis and must have a horizontal tangent at $x = 0$ .

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Example 2.30 (A quotient with an exponential). Find $f'(0)$ for $f(x) = \dfrac{e^x}{x^{2} + 1}$ .

Solution.

$f'(x) = \dfrac{e^x(x^{2}+1) - e^x\cdot 2x}{(x^{2}+1)^2} = \dfrac{e^x(x^{2} - 2x + 1)}{(x^{2}+1)^2} = \dfrac{e^x(x-1)^2}{(x^{2}+1)^2}$ .
At $x = 0$ : $f'(0) = \dfrac{1\cdot 1}{1} = 1$ .
Sanity check: $f'(x) \ge 0$ everywhere, with equality only at $x = 1$ , so the function is increasing and merely flattens momentarily at $x = 1$ .

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Pitfall. The numerator of the quotient rule is $f'g - fg'$ , in that order. Swapping the two terms changes the sign of every answer.

Exercises for §2.7 →

2.8Derivatives of the trigonometric functions

Throughout, angles are in radians; the formulas below are false in degrees. Two limits from the Limits chapter carry the whole section:

\lim_{h\to 0}\frac{\sin h}{h} = 1, \qquad \lim_{h\to 0}\frac{\cos h - 1}{h} = 0.

Theorem 2.31 (Derivatives of sine and cosine).

\frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x.

Proof. For the sine, use the addition formula $\sin(x+h) = \sin x\cos h + \cos x\sin h$ :

\frac{\sin(x+h) - \sin x}{h} = \frac{\sin x\cos h + \cos x\sin h - \sin x}{h} = \sin x\cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}.

As $h \to 0$ the first fraction tends to $0$ and the second to $1$ , so the limit is $\sin x\cdot 0 + \cos x\cdot 1 = \cos x$ .

For the cosine, use $\cos(x+h) = \cos x\cos h - \sin x\sin h$ :

\frac{\cos(x+h) - \cos x}{h} = \cos x\cdot\frac{\cos h - 1}{h} - \sin x\cdot\frac{\sin h}{h} \longrightarrow \cos x\cdot 0 - \sin x\cdot 1 = -\sin x.

∎

The two limits enter the proof in a way that explains them: $\frac{\sin h}{h} \to 1$ is the statement that $\sin$ has slope $1$ at $0$ , and $\frac{\cos h - 1}{h} \to 0$ is the statement that $\cos$ has a horizontal tangent at its peak. The addition formulas then transport those two facts at $0$ to every other point.

The other four trigonometric functions are quotients of these two, so the quotient rule finishes the job.

Theorem 2.32 (Derivatives of the other trigonometric functions).

\frac{d}{dx}\tan x = \sec^2 x, \qquad \frac{d}{dx}\sec x = \sec x\tan x, \qquad \frac{d}{dx}\cot x = -\csc^2 x, \qquad \frac{d}{dx}\csc x = -\csc x\cot x.

Proof. For $\tan x = \dfrac{\sin x}{\cos x}$ the quotient rule gives

\frac{\cos x\cdot\cos x - \sin x\cdot(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x.

For $\sec x = \dfrac{1}{\cos x}$ ,

\frac{0\cdot\cos x - 1\cdot(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x\tan x.

The cotangent and cosecant are the same two computations with the roles of sine and cosine exchanged, which is where their minus signs come from.∎

The list has a pattern that halves the memorising. Every co-function — cosine, cotangent, cosecant — carries a minus sign, and the formula for a co-function is obtained from the formula for its partner by swapping each function for its co-function and changing the sign.

Intuition. On the unit circle, a point moving anticlockwise at unit speed has coordinates $(\cos t, \sin t)$ . Its velocity vector is the position rotated a quarter turn, $(-\sin t, \cos t)$ , because velocity is always tangent to the circle. Read off the components: the rate of change of $\cos t$ is $-\sin t$ and that of $\sin t$ is $\cos t$ . The derivatives of sine and cosine are the geometry of uniform circular motion.

Example 2.33 (Simplifying before differentiating). Differentiate $y = \dfrac{\sin x}{1 + \cos x}$ .

Solution.

Quotient rule with $f = \sin x$ , $g = 1 + \cos x$ :

y' = \frac{\cos x(1+\cos x) - \sin x(-\sin x)}{(1+\cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2}.

Use $\cos^{2} x + \sin^{2} x = 1$ : the numerator is $1 + \cos x$ .
Cancel one factor: $y' = \dfrac{1}{1 + \cos x}$ .
Sanity check at $x = 0$ : $y'(0) = \frac12$ , and indeed near $0$ the function behaves like $\frac{x}{2}$ .

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Example 2.34 (A horizontal tangent). For $f(x) = \dfrac{\sec x}{1 + \tan x}$ , find $f'(x)$ and show that the tangent is horizontal at $x = \frac{\pi}{4}$ .

Solution.

Quotient rule with $f = \sec x$ , $g = 1 + \tan x$ , $f' = \sec x\tan x$ , $g' = \sec^{2} x$ :

f'(x) = \frac{\sec x\tan x(1 + \tan x) - \sec x\cdot\sec^2 x}{(1+\tan x)^2}.

Factor out $\sec x$ and use $\sec^{2} x = 1 + \tan^{2} x$ in the numerator: $\tan x + \tan^{2} x - 1 - \tan^{2} x = \tan x - 1$ . So

f'(x) = \frac{\sec x(\tan x - 1)}{(1 + \tan x)^2}.

At $x = \frac{\pi}{4}$ , $\tan x = 1$ , so the numerator vanishes while the denominator is $4$ : $f'(\frac{\pi}{4}) = 0$ , a horizontal tangent.

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Example 2.35 (An oscillating spring). A mass on a spring has position $s(t) = 4\cos t$ centimetres at time $t$ seconds. Find its velocity and acceleration, and the first time $t > 0$ at which it is momentarily at rest.

Solution.

Velocity is $v = s'(t) = -4\sin t$ and acceleration is $a = v'(t) = -4\cos t$ .
At rest means $v = 0$ , so $\sin t = 0$ ; the first positive solution is $t = \pi$ .
At that moment $s(\pi) = -4$ : the mass is at its lowest point, turning around. Notice $a = -s$ throughout, the defining property of simple harmonic motion.

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Pitfall. In degrees, $\frac{d}{dx}\sin x = \frac{\pi}{180}\cos x$ , because $x$ degrees is $\frac{\pi x}{180}$ radians. The clean formula $\sin' = \cos$ is the reason radians are the only unit used in calculus.

Exercises for §2.8 →

2.9Chain rule

Theorem 2.36 (Chain rule). If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , then the composite $F = f\circ g$ , $F(x) = f(g(x))$ , is differentiable at $x$ and

F'(x) = f'\!\left(g(x)\right)\cdot g'(x).

In Leibniz notation, with $y = f(u)$ and $u = g(x)$ ,

\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}.

Proof. Write $u = g(x)$ , let $\Delta x$ be a change in $x$ , and let $\Delta u = g(x+\Delta x) - g(x)$ and $\Delta y = f(u + \Delta u) - f(u)$ be the resulting changes. The tempting argument is

\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x},

and let $\Delta x \to 0$ . It is nearly right, but $\Delta u$ can be $0$ for small $\Delta x \neq 0$ (for instance when $g$ is constant near $x$ , or oscillates), and then $\frac{\Delta y}{\Delta u}$ is undefined. The repair is to avoid dividing by $\Delta u$ . Define

\varepsilon = \begin{cases} \dfrac{\Delta y}{\Delta u} - f'(u), & \Delta u \neq 0 \\[6pt] 0, & \Delta u = 0. \end{cases}

Then $\Delta y = f'(u)\,\Delta u + \varepsilon\,\Delta u$ holds in both cases: when $\Delta u \neq 0$ by the definition of $\varepsilon$ , and when $\Delta u = 0$ because both sides are $0$ . Moreover $\varepsilon \to 0$ as $\Delta u \to 0$ , by the definition of $f'(u)$ . Now divide by $\Delta x$ :

\frac{\Delta y}{\Delta x} = f'(u)\frac{\Delta u}{\Delta x} + \varepsilon\frac{\Delta u}{\Delta x}.

As $\Delta x \to 0$ , $\frac{\Delta u}{\Delta x} \to g'(x)$ , and $\Delta u \to 0$ because $g$ is continuous at $x$ , so $\varepsilon \to 0$ . The right side tends to $f'(u)g'(x) + 0\cdot g'(x) = f'(g(x))g'(x)$ .∎

In words: the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function. For $\frac{d}{dx}\sin(x^{3}) = \cos(x^{3})\cdot 3x^{2}$ the outer function is $\sin$ and the inner is $x^{3}$ . Compositions nest and so does the rule: peel one layer at a time, always evaluating each layer's derivative at whatever is inside it.

This is the rule that matters most. Implicit differentiation, related rates, and integration by substitution are all the chain rule wearing different clothes, and a missing inner derivative is the most common mistake in the whole of calculus. Three special cases are used so often they deserve their own lines.

Corollary 2.37 (Chain rule with powers and exponentials). If $g$ is differentiable, then

\frac{d}{dx}\left[g(x)\right]^n = n\left[g(x)\right]^{n-1}g'(x), \qquad \frac{d}{dx}e^{g(x)} = e^{g(x)}g'(x), \qquad \frac{d}{dx}e^{kx} = ke^{kx}.

The chain rule also settles the constant left hanging in the exponential section.

Theorem 2.38 (Derivative of $a^x$ ). For $a > 0$ , $\dfrac{d}{dx}a^x = a^x\ln a$ . Consequently $\displaystyle\lim_{h\to 0}\frac{a^h - 1}{h} = \ln a$ .

Proof. Since $a = e^{\ln a}$ , we have $a^x = \left(e^{\ln a}\right)^x = e^{x\ln a}$ . This is $e^{g(x)}$ with $g(x) = x\ln a$ and $g'(x) = \ln a$ , so by the chain rule

\frac{d}{dx}a^x = e^{x\ln a}\cdot\ln a = a^x\ln a.

Comparing with $\frac{d}{dx}a^x = L(a)\,a^x$ from the exponential section gives $L(a) = \ln a$ .∎

Intuition. Picture two connected gears: the big one turns the small one. To know how fast the output turns you need both ratios, and you multiply them.

Concretely: a $1^{\circ}$ rise in temperature sells $50$ more ice creams, and each earns $3$ dollars, so revenue rises by $50\times 3 = 150$ dollars per degree. Two rates, multiplied — that is the chain rule, and it is why Leibniz's $\frac{dy}{du}\cdot\frac{du}{dx}$ looks like cancelling fractions.

Example 2.39 (A power of a polynomial). Find $\frac{d}{dx}\left(x^{3} + 1\right)^{100}$ .

Solution.

Outer function $u^{100}$ , inner function $u = x^{3} + 1$ .
Derivative of the outer at the inner: $100(x^{3} + 1)^{99}$ . Derivative of the inner: $3x^{2}$ .
Multiply: $300x^{2}(x^{3}+1)^{99}$ .
Sanity check: expanding $(x^{3}+1)^{100}$ would produce a polynomial of degree $300$ with leading term $x^{300}$ , whose derivative starts $300x^{299}$ , and so does ours.

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Example 2.40 (Chain rule inside a product). Find $\frac{d}{dx}\left[xe^{-x^{2}}\right]$ .

Solution.

This is a product of $x$ and $e^{-x^{2}}$ ; the second factor needs the chain rule.
$\frac{d}{dx}e^{-x^{2}} = e^{-x^{2}}\cdot(-2x)$ , by the exponential case of the corollary with $g(x) = -x^{2}$ .
Product rule: $1\cdot e^{-x^{2}} + x\cdot(-2x)e^{-x^{2}} = e^{-x^{2}}(1 - 2x^{2})$ .
Sanity check: the derivative vanishes at $x = \pm\frac{1}{\sqrt2}$ , where the bell-shaped graph of $xe^{-x^{2}}$ has its peak and trough.

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Example 2.41 (Three nested layers). Find $\frac{d}{dx}e^{\sin(x^{2})}$ .

Solution.

The layers are $e^{(\cdot)}$ , then $\sin(\cdot)$ , then $x^{2}$ ; peel them from the outside in.
Outermost: $\frac{d}{du}e^u = e^u$ , which at $u = \sin(x^{2})$ is $e^{\sin(x^{2})}$ .
Middle: $\frac{d}{dv}\sin v = \cos v$ , which at $v = x^{2}$ is $\cos(x^{2})$ .
Innermost: $\frac{d}{dx}x^{2} = 2x$ .
Multiply all three: $\frac{d}{dx}e^{\sin(x^{2})} = 2x\cos(x^{2})\,e^{\sin(x^{2})}$ .

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Example 2.42 (A composite from a table). Suppose $g(2) = 5$ , $g'(2) = -3$ , $f'(5) = 4$ and $f'(2) = 7$ . Find $(f\circ g)'(2)$ .

Solution.

By the chain rule $(f\circ g)'(2) = f'(g(2))\cdot g'(2)$ .
The outer derivative is evaluated at the inner value $g(2) = 5$ , so we need $f'(5) = 4$ , not $f'(2)$ .
$(f\circ g)'(2) = 4\cdot(-3) = -12$ .
The value $f'(2) = 7$ is a decoy: it is the derivative of $f$ at the wrong point.

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Example 2.43 (Chain, quotient and power together). Differentiate $y = \dfrac{(2x+1)^5}{(x^{3} - x + 1)^4}$ and simplify.

Solution.

Set $f = (2x+1)^5$ and $g = (x^{3} - x + 1)^4$ . By the power case of the chain rule, $f' = 5(2x+1)^4\cdot 2 = 10(2x+1)^4$ and $g' = 4(x^{3} - x + 1)^3(3x^{2} - 1)$ .
Quotient rule:

y' = \frac{10(2x+1)^4(x^3-x+1)^4 - (2x+1)^5\cdot 4(x^3-x+1)^3(3x^2-1)}{(x^3-x+1)^8}.

Both terms of the numerator contain $2(2x+1)^4(x^{3}-x+1)^3$ ; factor it out and cancel three powers of the denominator:

y' = \frac{2(2x+1)^4\left[5(x^3-x+1) - 2(2x+1)(3x^2-1)\right]}{(x^3-x+1)^5}.

Expand the bracket: $5x^{3} - 5x + 5 - (12x^{3} + 6x^{2} - 4x - 2) = -7x^{3} - 6x^{2} - x + 7$ . So

y' = \frac{2(2x+1)^4\left(-7x^3 - 6x^2 - x + 7\right)}{(x^3-x+1)^5}.

Sanity check at $x = 0$ : directly, $f = 1$ , $g = 1$ , $f' = 10$ , $g' = -4$ , so $y'(0) = \frac{10\cdot 1 - 1\cdot(-4)}{1} = 14$ ; the simplified formula gives $\frac{2\cdot 1\cdot 7}{1} = 14$ .

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Pitfall. $\sin^{2} x$ and $\sin(x^{2})$ are different functions with different derivatives: the first is $(\sin x)^2$ , with derivative $2\sin x\cos x$ ; the second is $\sin$ of $x^{2}$ , with derivative $2x\cos(x^{2})$ . Decide which layer is outermost before you differentiate anything.

Exercises for §2.9 →

2.10Implicit differentiation

Some curves refuse to be written as $y = f(x)$ . A circle $x^{2} + y^{2} = 25$ fails the vertical line test, yet it has a perfectly good tangent line at every point except the two where the tangent is vertical. Implicit differentiation finds that slope without ever solving for $y$ .

Method 2.44 (Implicit differentiation). Given an equation in $x$ and $y$ that defines $y$ as a function of $x$ near a point:

Differentiate both sides with respect to $x$ , treating $y$ as an unknown function $y(x)$ . Every term containing $y$ picks up a factor $\frac{dy}{dx}$ through the chain rule.
Collect the terms containing $\frac{dy}{dx}$ on one side and everything else on the other.
Solve for $\frac{dy}{dx}$ . The answer will usually involve both $x$ and $y$ .

The method is the chain rule and nothing else. If $y$ is a function of $x$ , then $y^{2}$ is the composite $[y(x)]^2$ , whose derivative is $2y\cdot y'$ ; likewise $\frac{d}{dx}\sin y = \cos y\cdot y'$ and $\frac{d}{dx}(xy) = y + xy'$ by the product rule. That the equation defines $y$ as a differentiable function near the point is an assumption; the Implicit Function Theorem of multivariable calculus says it holds whenever the partial derivative with respect to $y$ is nonzero there, which in practice means the denominator of your answer is not zero.

Intuition. Usually $y$ sits alone on one side, as in $y = x^{2} + 3$ . Sometimes $x$ and $y$ are tangled together and cannot be separated.

Differentiate anyway. Every time you meet a $y$ , remember it secretly depends on $x$ and attach $\frac{dy}{dx}$ ; then solve for that symbol. It is detective work: you learn how fast the hidden variable changes from the equation that traps it.

Example 2.45 (Implicit differentiation of a circle). Find $\frac{dy}{dx}$ for $x^{2} + y^{2} = 25$ , and evaluate it at $(3,4)$ .

Solution.

Differentiate both sides with respect to $x$ : the left side gives $2x + 2y\frac{dy}{dx}$ , where the second term is the chain rule applied to $y^{2}$ ; the right side gives $0$ .
So $2x + 2y\frac{dy}{dx} = 0$ , hence $\frac{dy}{dx} = -\frac{x}{y}$ .
At $(3,4)$ the slope is $-\frac34$ .
Sanity check: the radius to $(3,4)$ has slope $\frac43$ , and the tangent to a circle is perpendicular to the radius; $-\frac34$ is the negative reciprocal.

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Example 2.46 (The folium of Descartes). The curve $x^{3} + y^{3} = 6xy$ is the folium of Descartes. Find $\frac{dy}{dx}$ , the tangent line at $(3,3)$ , and the point in the first quadrant where the tangent is horizontal.

Solution.

Differentiate: $3x^{2} + 3y^{2} y' = 6y + 6xy'$ , using the product rule on $6xy$ .
Collect: $3y^{2} y' - 6xy' = 6y - 3x^{2}$ , so $y'(y^{2} - 2x) = 2y - x^{2}$ and

\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}.

At $(3,3)$ : $y' = \dfrac{6 - 9}{9 - 6} = -1$ . The tangent line is $y - 3 = -(x - 3)$ , that is $x + y = 6$ .
Horizontal tangent means the numerator vanishes: $2y = x^{2}$ , so $y = \frac{x^{2}}{2}$ . Substitute into the curve: $x^{3} + \frac{x^{6}}{8} = 3x^{3}$ , so $x^{6} = 16x^{3}$ and, for $x \neq 0$ , $x^{3} = 16$ , $x = 2^{4/3}$ . Then $y = \frac{x^{2}}{2} = 2^{8/3 - 1} = 2^{5/3}$ .
The point is $\left(2^{4/3}, 2^{5/3}\right) \approx (2.52, 3.17)$ . Sanity check: it lies on the loop above the line $y = x$ , where the loop's top must be.

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Example 2.47 (Sine and cosine of both variables). Find $y'$ if $\sin(x + y) = y^{2}\cos x$ .

Solution.

Left side, chain rule: $\cos(x+y)\cdot(1 + y')$ .
Right side, product rule with the chain rule on $y^{2}$ : $2yy'\cos x - y^{2}\sin x$ .
Equate the two sides and expand the left:

\cos(x+y) + \cos(x+y)\,y' = 2y\cos x\,y' - y^2\sin x.

Collect the $y'$ terms on the left and everything else on the right: $y'\left[\cos(x+y) - 2y\cos x\right] = -y^{2}\sin x - \cos(x+y)$ .
Solve, multiplying top and bottom by $-1$ to tidy the signs: $y' = \dfrac{y^{2}\sin x + \cos(x+y)}{2y\cos x - \cos(x+y)}$ .

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Example 2.48 (A second derivative, implicitly). Find $y''$ for the circle $x^{2} + y^{2} = 25$ .

Solution.

From the first example, $y' = -\dfrac{x}{y}$ .
Differentiate this with the quotient rule, remembering $y$ depends on $x$ :

y'' = -\frac{1\cdot y - x\,y'}{y^2}.

Substitute $y' = -x/y$ to remove the derivative:

y'' = -\frac{y - x(-x/y)}{y^2} = -\frac{y + x^2/y}{y^2} = -\frac{y^2 + x^2}{y^3}.

Use the original equation $x^{2} + y^{2} = 25$ to simplify: $y'' = -\dfrac{25}{y^{3}}$ .
Sanity check: on the upper semicircle $y > 0$ , so $y'' < 0$ and the arc is concave down, as a dome should be.

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Pitfall. Every derivative of a $y$ -term carries a factor $\frac{dy}{dx}$ . Writing $\frac{d}{dx}y^{2} = 2y$ instead of $2y\frac{dy}{dx}$ is the standard error and produces an answer with no $\frac{dy}{dx}$ to solve for.

Exercises for §2.10 →

2.11Derivatives of inverse functions and the logarithm

If $f$ is one-to-one it has an inverse $f^{-1}$ , whose graph is the reflection of the graph of $f$ in the line $y = x$ . Reflection swaps rise and run, so it should invert slopes; the theorem below says exactly that, with the one caveat that a horizontal tangent on $f$ reflects to a vertical tangent on $f^{-1}$ , where there is no slope.

Theorem 2.49 (Derivative of an inverse function). Let $f$ be one-to-one and continuous on an interval, and differentiable at $b = f^{-1}(a)$ with $f'(b) \neq 0$ . Then $f^{-1}$ is differentiable at $a$ and

\left(f^{-1}\right)'(a) = \frac{1}{f'\!\left(f^{-1}(a)\right)}.

In Leibniz notation, $\dfrac{dx}{dy} = \dfrac{1}{dy/dx}$ .

Proof. Write $g = f^{-1}$ . For $x \neq a$ in the range of $f$ , put $y = g(x)$ and $b = g(a)$ ; then $y \neq b$ because $g$ is one-to-one, and $f(y) = x$ , $f(b) = a$ . So

\frac{g(x) - g(a)}{x - a} = \frac{y - b}{f(y) - f(b)} = \frac{1}{\dfrac{f(y)-f(b)}{y-b}}.

A continuous one-to-one function on an interval has a continuous inverse — this is a fact from real analysis, and it is the only step not proved here — so as $x \to a$ we have $y = g(x) \to g(a) = b$ . The inner difference quotient therefore tends to $f'(b) \neq 0$ , and the whole expression tends to $\dfrac{1}{f'(b)}$ .∎

There is a quicker route once differentiability of $f^{-1}$ is known: differentiate the identity $f(f^{-1}(x)) = x$ with the chain rule to get $f'(f^{-1}(x))\cdot(f^{-1})'(x) = 1$ and divide. That argument assumes the inverse is differentiable; the proof above establishes it. In practice the quick route, or simply implicit differentiation, is how the formulas for specific inverses are found.

The most important inverse function is the natural logarithm, the inverse of $e^x$ .

Theorem 2.50 (Derivative of the natural logarithm). For $x > 0$ ,

\frac{d}{dx}\ln x = \frac{1}{x}.

More generally, for $x \neq 0$ , $\dfrac{d}{dx}\ln|x| = \dfrac{1}{x}$ , and for any base $a > 0$ , $a \neq 1$ ,

\frac{d}{dx}\log_a x = \frac{1}{x\ln a}.

Proof. Let $y = \ln x$ , so that $e^y = x$ . Differentiate both sides implicitly with respect to $x$ : $e^y\,y' = 1$ , so

y' = \frac{1}{e^y} = \frac{1}{x}.

For $x < 0$ , $\ln|x| = \ln(-x)$ , and the chain rule gives $\dfrac{1}{-x}\cdot(-1) = \dfrac{1}{x}$ . For a general base, $\log_a x = \dfrac{\ln x}{\ln a}$ by the change-of-base formula, and $\ln a$ is a constant.∎

The formula is remarkable: the logarithm, a transcendental function, has for its derivative the simplest rational function there is. This is why $\frac{1}{x}$ , the one power of $x$ the power rule cannot produce as a derivative (it would need $x^{0}/0$ ), has $\ln|x|$ for its antiderivative. Combined with the chain rule,

\frac{d}{dx}\ln\left(g(x)\right) = \frac{g'(x)}{g(x)},

a ratio worth recognising on sight: the derivative of the inside over the inside.

Intuition. Reflect a graph in the line $y = x$ and every rise becomes a run. A tangent that climbs $3$ units for every $1$ unit across, slope $3$ , reflects to one that climbs $1$ for every $3$ , slope $\frac13$ . For $e^x$ at the point $(1, e)$ the slope is $e$ ; so at the reflected point $(e, 1)$ the slope of $\ln x$ is $\frac1e$ , and indeed $\frac{1}{x}$ at $x = e$ is $\frac1e$ .

Example 2.51 (The derivative of an inverse at a point). Let $f(x) = x^{3} + x$ . Find $\left(f^{-1}\right)'(2)$ .

Solution.

$f$ is increasing (since $f'(x) = 3x^{2} + 1 > 0$ ), so it is one-to-one and has an inverse.
We need $f^{-1}(2)$ : solve $x^{3} + x = 2$ , and $x = 1$ works, so $f^{-1}(2) = 1$ .
$f'(1) = 3 + 1 = 4$ .
By the theorem, $\left(f^{-1}\right)'(2) = \dfrac{1}{f'(1)} = \dfrac14$ . Solving for $f^{-1}$ explicitly would require the cubic formula; the theorem sidesteps it.

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Example 2.52 (Logarithm laws before the chain rule). Differentiate $y = \ln\dfrac{x+1}{\sqrt{x-2}}$ for $x > 2$ .

Solution.

Expand with the logarithm laws first: $y = \ln(x+1) - \tfrac12\ln(x-2)$ .
Each term is now $\ln$ of something simple: $y' = \dfrac{1}{x+1} - \dfrac{1}{2(x-2)}$ .
Differentiating the quotient inside the logarithm directly gives the same answer after considerably more algebra. Expand logarithms first whenever you can.

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Example 2.53 (A logarithm to another base). Find $\frac{d}{dx}\log_2(3x + 1)$ .

Solution.

Write $\log_2 u = \dfrac{\ln u}{\ln 2}$ with $u = 3x + 1$ .
$\frac{d}{dx}\ln(3x+1) = \dfrac{3}{3x+1}$ by the chain rule.
So the derivative is $\dfrac{3}{(3x+1)\ln 2}$ .

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Example 2.54 (Recognising a derivative). Show that $F(x) = x\ln x - x$ satisfies $F'(x) = \ln x$ .

Solution.

Product rule on $x\ln x$ : $1\cdot\ln x + x\cdot\dfrac{1}{x} = \ln x + 1$ .
The derivative of $-x$ is $-1$ .
So $F'(x) = \ln x + 1 - 1 = \ln x$ . In the language of the next chapters, $x\ln x - x$ is an antiderivative of $\ln x$ .

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Pitfall. $\left(f^{-1}\right)'(x)$ is $\dfrac{1}{f'(f^{-1}(x))}$ , not $\dfrac{1}{f'(x)}$ . The reciprocal slope must be taken at the corresponding point on the other graph. For $f(x) = x^{3} + x$ above, $\frac{1}{f'(2)} = \frac{1}{13}$ is wrong; the answer is $\frac{1}{f'(1)} = \frac14$ .

Exercises for §2.11 →

2.12Inverse trigonometric derivatives

Theorem 2.55 (Derivatives of the inverse trigonometric functions).

\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\arctan x = \frac{1}{1+x^2},

the first two for $|x| < 1$ and the third for all $x$ . Also $\dfrac{d}{dx}\operatorname{arcsec} x = \dfrac{1}{x\sqrt{x^{2}-1}}$ for $|x| > 1$ , while $\operatorname{arccot}$ and $\operatorname{arccsc}$ have the negatives of the derivatives of $\arctan$ and $\operatorname{arcsec}$ .

Proof. Let $y = \arcsin x$ , so $\sin y = x$ with $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ . Differentiate implicitly: $\cos y\cdot y' = 1$ , so $y' = \dfrac{1}{\cos y}$ . On this range of $y$ , $\cos y \ge 0$ , so $\cos y = +\sqrt{1 - \sin^{2} y} = \sqrt{1 - x^{2}}$ , and

\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}.

The choice of sign for the square root is exactly where the restriction of $\arcsin$ to $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is used; at the endpoints $\cos y = 0$ and the derivative is infinite, matching the vertical tangents of $\arcsin$ at $x = \pm 1$ .

For $y = \arctan x$ : $\tan y = x$ , so $\sec^{2} y\cdot y' = 1$ and $y' = \dfrac{1}{\sec^{2} y} = \dfrac{1}{1 + \tan^{2} y} = \dfrac{1}{1+x^{2}}$ . No square root and no sign question arise, because $\sec^{2} y$ is a polynomial in $\tan y$ .

For $\arccos$ , either repeat the argument with $\sin y \ge 0$ on $[0, \pi]$ , or use $\arccos x = \frac{\pi}{2} - \arcsin x$ and differentiate. The remaining three follow the same pattern.∎

That $\arcsin$ and $\arccos$ have opposite derivatives is no accident: their sum is the constant $\frac{\pi}{2}$ , so the derivatives must cancel. The arctangent is the one you will meet most often, because $\arctan x + C$ is the antiderivative of $\frac{1}{1+x^{2}}$ . Composed with the chain rule it reads

\frac{d}{dx}\arctan\left(g(x)\right) = \frac{g'(x)}{1 + [g(x)]^2},

so for example $\frac{d}{dx}\arctan(3x) = \frac{3}{1+9x^{2}}$ .

Intuition. An inverse trigonometric function answers "which angle has this sine?", and its derivative says how sensitive that angle is to a small change in the value. Near $x = 0$ , $\arcsin$ has slope $1$ : sine and angle are nearly interchangeable for small angles. Near $x = 1$ the slope is enormous, because the sine barely moves as the angle approaches $\frac{\pi}{2}$ , so recovering the angle from the sine is extremely sensitive there.

The square roots come from the Pythagorean theorem: draw a right triangle with opposite side $x$ and hypotenuse $1$ , and the adjacent side, $\sqrt{1-x^{2}}$ , is exactly the denominator.

Example 2.56 (Inverse trigonometric derivative with the chain rule). Find $\frac{d}{dx}\arcsin(x^{2})$ .

Solution.

Outer function $\arcsin u$ with $u = x^{2}$ .
Derivative of the outer at the inner: $\dfrac{1}{\sqrt{1-u^{2}}} = \dfrac{1}{\sqrt{1-x^{4}}}$ . Derivative of the inner: $2x$ .
Multiply: $\dfrac{2x}{\sqrt{1-x^{4}}}$ , valid for $|x| < 1$ .

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Example 2.57 (Arctangent of a root). Find $\frac{d}{dx}\arctan\sqrt{x}$ for $x > 0$ .

Solution.

Outer $\arctan u$ , inner $u = \sqrt{x}$ with $u' = \dfrac{1}{2\sqrt{x}}$ .
$\dfrac{1}{1 + u^{2}}\cdot u' = \dfrac{1}{1 + x}\cdot\dfrac{1}{2\sqrt{x}} = \dfrac{1}{2\sqrt{x}\,(1+x)}$ .

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Example 2.58 (A surprising simplification). Show that $\dfrac{d}{dx}\arctan\dfrac{1+x}{1-x} = \dfrac{1}{1+x^{2}}$ for $x \neq 1$ .

Solution.

Let $u = \dfrac{1+x}{1-x}$ . By the quotient rule, $u' = \dfrac{(1)(1-x) - (1+x)(-1)}{(1-x)^2} = \dfrac{2}{(1-x)^2}$ .
Compute $1 + u^{2} = \dfrac{(1-x)^2 + (1+x)^2}{(1-x)^2} = \dfrac{2 + 2x^{2}}{(1-x)^2}$ .
Chain rule: $\dfrac{u'}{1+u^{2}} = \dfrac{2}{(1-x)^2}\cdot\dfrac{(1-x)^2}{2(1+x^{2})} = \dfrac{1}{1+x^{2}}$ .
The derivative equals that of $\arctan x$ itself, so the two functions differ by a constant on each side of $x = 1$ ; for $x < 1$ that constant is $\frac{\pi}{4}$ , as substituting $x = 0$ shows. This is the tangent addition formula seen through calculus.

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Pitfall. The arcsine derivative has a square root, the arctangent derivative does not: $\frac{1}{\sqrt{1-x^{2}}}$ versus $\frac{1}{1+x^{2}}$ . Mixing the two, or writing $1 - x^{2}$ under the arctangent, is a frequent slip. Note also that $\frac{d}{dx}\arcsin x$ is only defined for $|x| < 1$ .

Exercises for §2.12 →

2.13Logarithmic differentiation

When the variable appears in both the base and the exponent, as in $x^x$ , neither the power rule nor the exponential rule applies: the power rule needs a constant exponent, the exponential rule a constant base. Taking logarithms first repairs this, because $\ln(f^g) = g\ln f$ pulls the exponent down into a product. The same move turns products into sums and quotients into differences, which is why the method also tames long products and quotients with constant exponents.

Method 2.59 (Logarithmic differentiation). Let $y = f(x)$ with $f(x) > 0$ (otherwise use $|y|$ ).

Take the natural logarithm of both sides and expand with the logarithm laws.
Differentiate implicitly with respect to $x$ . The left side becomes $\dfrac{1}{y}\dfrac{dy}{dx}$ .
Multiply through by $y$ , replacing $y$ by the original expression.

The method pays the final instalment on the power rule.

Theorem 2.60 (Power rule for real exponents). For every real number $n$ and all $x > 0$ , $\dfrac{d}{dx}x^n = nx^{n-1}$ . The formula also holds for $x < 0$ whenever $x^n$ is defined there, and at $x = 0$ when $n \ge 1$ and $x^n$ is defined on both sides of $0$ .

Proof. For $x > 0$ let $y = x^n$ , so $\ln y = n\ln x$ . Differentiating, $\dfrac{y'}{y} = \dfrac{n}{x}$ , hence

y' = \frac{ny}{x} = \frac{nx^n}{x} = nx^{n-1}.

For $x < 0$ with $x^n$ defined (for instance $n$ a rational with odd denominator), apply the same argument to $|y| = |x|^n$ : $\frac{y'}{y} = \frac{n}{x}$ again, since $\frac{d}{dx}\ln|x| = \frac1x$ . The case $x = 0$ is the definition directly.∎

Intuition. Some expressions are nested puzzles where no single rule fits. The logarithm takes the puzzle apart into pieces you can differentiate, and multiplying by $y$ at the end puts it back together. It is also a bookkeeping device: the relative rate of change $\frac{y'}{y}$ of a product is the sum of the relative rates of its factors, which is the product rule in a form that scales to any number of factors.

Example 2.61 (Differentiating $x^x$ ). Find $\frac{d}{dx}x^x$ for $x > 0$ .

Solution.

Set $y = x^x$ and take logarithms: $\ln y = x\ln x$ .
Differentiate both sides. The left gives $\dfrac{1}{y}\dfrac{dy}{dx}$ by the chain rule.
The right needs the product rule: $\frac{d}{dx}(x\ln x) = \ln x + x\cdot\frac{1}{x} = \ln x + 1$ .
So $\dfrac{1}{y}\dfrac{dy}{dx} = \ln x + 1$ , and multiplying by $y = x^x$ gives $\dfrac{dy}{dx} = x^x(\ln x + 1)$ .
The derivative vanishes when $\ln x = -1$ , that is at $x = 1/e$ — the minimum of $x^x$ . Note also that $x^x = e^{x\ln x}$ gives the same answer through the chain rule; the two methods are the same computation.

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Example 2.62 (A long product and quotient). Differentiate $y = \dfrac{x^{3/4}\sqrt{x^{2}+1}}{(3x+2)^5}$ for $x > 0$ .

Solution.

Take logarithms and expand: $\ln y = \tfrac34\ln x + \tfrac12\ln(x^{2}+1) - 5\ln(3x+2)$ .
Differentiate: $\dfrac{y'}{y} = \dfrac{3}{4x} + \dfrac{x}{x^{2}+1} - \dfrac{15}{3x+2}$ .
Multiply by $y$ :

y' = \frac{x^{3/4}\sqrt{x^2+1}}{(3x+2)^5}\left(\frac{3}{4x} + \frac{x}{x^2+1} - \frac{15}{3x+2}\right).

The product and quotient rules would give the same answer as a single fraction after a page of algebra; here each factor contributed one easy term.

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Example 2.63 (A variable base and exponent). Find $\frac{d}{dx}\left(1 + \dfrac{1}{x}\right)^x$ for $x > 0$ .

Solution.

Let $y = \left(1 + \frac1x\right)^x$ , so $\ln y = x\ln\left(1 + \frac1x\right)$ .
Product rule and chain rule:

\frac{y'}{y} = \ln\left(1 + \frac1x\right) + x\cdot\frac{1}{1 + \frac1x}\cdot\left(-\frac{1}{x^2}\right) = \ln\left(1 + \frac1x\right) - \frac{1}{x + 1}.

So $y' = \left(1 + \dfrac1x\right)^x\left[\ln\left(1 + \dfrac1x\right) - \dfrac{1}{x+1}\right]$ .
Sanity check: as $x \to \infty$ the bracket tends to $0$ and $y \to e$ ; the function levels off at its famous limit.

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Pitfall. Logarithms of negative numbers do not exist, so $\ln y$ only makes sense where $y > 0$ . When $y$ can be negative, take $\ln|y|$ instead: the derivative of $\ln|y|$ is still $\frac{y'}{y}$ , so the recipe is unchanged.

Exercises for §2.13 →

2.14The table of derivatives

Every derivative you will ever compute reduces, through the product, quotient and chain rules, to the following list. It is the vocabulary of the subject and has to be known by heart; every entry has been proved earlier in this chapter or in the hyperbolic section at its end.

Summary (Derivatives of the standard functions). Powers and exponentials:

\frac{d}{dx}x^n = nx^{n-1}, \qquad \frac{d}{dx}e^x = e^x, \qquad \frac{d}{dx}a^x = a^x\ln a.

Logarithms:

\frac{d}{dx}\ln|x| = \frac{1}{x}, \qquad \frac{d}{dx}\log_a x = \frac{1}{x\ln a}.

Trigonometric:

\frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x, \qquad \frac{d}{dx}\tan x = \sec^2 x,

\frac{d}{dx}\sec x = \sec x\tan x, \qquad \frac{d}{dx}\csc x = -\csc x\cot x, \qquad \frac{d}{dx}\cot x = -\csc^2 x.

Inverse trigonometric:

\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\arctan x = \frac{1}{1+x^2}.

Hyperbolic:

\frac{d}{dx}\sinh x = \cosh x, \qquad \frac{d}{dx}\cosh x = \sinh x, \qquad \frac{d}{dx}\tanh x = \operatorname{sech}^2 x.

Combining rules: $(cf)' = cf'$ , $(f \pm g)' = f' \pm g'$ , $(fg)' = f'g + fg'$ , $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^{2}}$ , and $\left(f\circ g\right)' = \left(f'\circ g\right)\cdot g'$ .

Two patterns make the list shorter than it looks. Among the exponentials and logarithms, $e$ is the base that needs no correction, and every other base pays a factor of $\ln a$ — multiplied for $a^x$ , divided for $\log_a x$ . Among the trigonometric functions, every co-function carries a minus sign, and the sine family cycles $\sin \to \cos \to -\sin \to -\cos \to \sin$ . The hyperbolic family has the same cycle with the minus signs removed.

Intuition. Think of these as vocabulary words. You cannot write sentences without words, and you cannot differentiate without these; the product, quotient and chain rules are only the grammar that combines them. A derivative problem is a parsing problem: identify the outermost operation (sum? product? quotient? composition?), apply its rule, and repeat on the pieces until every piece is in the table.

Example 2.64 (A mixed sum). Differentiate $f(x) = 4\sec x - 2\ln x + 3^x + \arctan x$ .

Solution.

It is a sum, so work term by term from the table.
$\frac{d}{dx}(4\sec x) = 4\sec x\tan x$ ; $\frac{d}{dx}(-2\ln x) = -\dfrac{2}{x}$ ; $\frac{d}{dx}3^x = 3^x\ln 3$ ; $\frac{d}{dx}\arctan x = \dfrac{1}{1+x^{2}}$ .
$f'(x) = 4\sec x\tan x - \dfrac{2}{x} + 3^x\ln 3 + \dfrac{1}{1+x^{2}}$ .

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Example 2.65 (Fitting initial conditions). Find constants $A$ and $B$ so that $y = A\cos x + B\sin x$ satisfies $y(0) = 2$ and $y'(0) = -3$ .

Solution.

$y(0) = A\cos 0 + B\sin 0 = A$ , so $A = 2$ .
$y' = -A\sin x + B\cos x$ , so $y'(0) = B$ and $B = -3$ .
The function is $y = 2\cos x - 3\sin x$ . It also satisfies $y'' = -y$ , a first taste of how derivatives specify functions through differential equations.

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Example 2.66 (Evaluating a derivative at a point). Let $g(x) = \dfrac{\ln x}{x^{2}}$ . Find $g'(e)$ .

Solution.

Quotient rule: $g'(x) = \dfrac{\frac1x\cdot x^{2} - \ln x\cdot 2x}{x^{4}} = \dfrac{x - 2x\ln x}{x^{4}} = \dfrac{1 - 2\ln x}{x^{3}}$ .
At $x = e$ , $\ln e = 1$ : $g'(e) = \dfrac{1 - 2}{e^{3}} = -\dfrac{1}{e^{3}}$ .
Sanity check: $g'(x) = 0$ at $\ln x = \frac12$ , i.e. $x = \sqrt{e} < e$ , so at $x = e$ the function is already past its peak and decreasing; the negative sign is right.

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Exercises for §2.14 →

2.15Higher-order derivatives

Definition 2.67 (Second and higher derivatives). The second derivative $f''$ is the derivative of $f'$ , and in general the $n$ th derivative $f^{(n)}$ is the derivative of $f^{(n-1)}$ . Equivalent notations are

f''(x) = y'' = \frac{d^2y}{dx^2} = D^2 f(x), \qquad f^{(n)}(x) = y^{(n)} = \frac{d^ny}{dx^n} = D^n f(x).

The second derivative is the rate at which the rate is changing. If $s(t)$ is position then $s'(t)$ is velocity and $s''(t)$ is acceleration: positive under the accelerator, negative under the brake. (The third derivative even has a name, jerk — how abruptly the acceleration changes.)

Geometrically, $f'' > 0$ means $f'$ is increasing, so the graph of $f$ bends upward like a bowl — concave up; $f'' < 0$ means it bends downward like a hill. This is what the second derivative test for extrema reads, in the applications chapter.

Higher derivatives are also what a Taylor series is made of: the $n$ th coefficient at $a$ is $\frac{f^{(n)}(a)}{n!}$ , which is how a smooth function is approximated by a polynomial. Functions with periodic patterns in their higher derivatives, such as $e^x$ , $\sin x$ and $\frac1x$ , therefore have especially clean series, and finding the pattern is a standard exercise.

Intuition. In a car, position is what the odometer reads, velocity is what the speedometer reads, and acceleration is what you feel pressing you into the seat. Each is the derivative of the one before. Constant velocity means zero acceleration: on a straight motorway at steady speed you feel nothing, even though the odometer is climbing steadily.

Example 2.68 (Successive derivatives of a polynomial). Find all the derivatives of $f(x) = x^{4} - 3x^{2} + 5x$ .

Solution.

$f'(x) = 4x^{3} - 6x + 5$ and $f''(x) = 12x^{2} - 6$ .
$f'''(x) = 24x$ and $f^{(4)}(x) = 24$ .
$f^{(n)}(x) = 0$ for all $n \ge 5$ : each derivative lowers the degree by one, so a polynomial of degree $n$ has $f^{(n)}$ constant and everything beyond that zero.

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Example 2.69 (A pattern in the derivatives of $1/x$ ). Find a formula for $f^{(n)}(x)$ when $f(x) = \dfrac{1}{x}$ .

Solution.

Write $f(x) = x^{-1}$ and differentiate repeatedly: $f' = -x^{-2}$ , $f'' = 2x^{-3}$ , $f''' = -6x^{-4}$ , $f^{(4)} = 24x^{-5}$ .
The signs alternate, the coefficients are $1, 2, 6, 24 = 1!, 2!, 3!, 4!$ , and the exponent is $-(n+1)$ .
So $f^{(n)}(x) = \dfrac{(-1)^n\,n!}{x^{n+1}}$ . Check with $n = 3$ : $\frac{(-1)^3 3!}{x^{4}} = -\frac{6}{x^{4}}$ , as computed.

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Example 2.70 (The 27th derivative of cosine). Find $\dfrac{d^{27}}{dx^{27}}\cos x$ .

Solution.

The derivatives of $\cos x$ cycle with period $4$ : $\cos x \to -\sin x \to -\cos x \to \sin x \to \cos x$ .
Since $27 = 24 + 3$ , the 27th derivative equals the third derivative.
The third derivative of $\cos x$ is $\sin x$ .

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Example 2.71 (Verifying a differential equation). Show that $y = e^{-x}\cos x$ satisfies $y'' + 2y' + 2y = 0$ .

Solution.

Product rule: $y' = -e^{-x}\cos x - e^{-x}\sin x = -e^{-x}(\cos x + \sin x)$ .
Again: $y'' = e^{-x}(\cos x + \sin x) - e^{-x}(-\sin x + \cos x) = e^{-x}\cdot 2\sin x = 2e^{-x}\sin x$ .
Substitute:

y'' + 2y' + 2y = 2e^{-x}\sin x - 2e^{-x}(\cos x + \sin x) + 2e^{-x}\cos x = 0.

Every term cancels. This is a damped oscillation: the $e^{-x}$ factor shrinks the amplitude while $\cos x$ supplies the wobble.

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Pitfall. $f''(x)$ is the derivative of $f'(x)$ , not its square. For $f(x) = x^{3}$ , $f''(x) = 6x$ while $[f'(x)]^2 = 9x^{4}$ . The Leibniz notation $\frac{d^{2}y}{dx^{2}}$ keeps the distinction visible.

Exercises for §2.15 →

2.16Tangent lines, linear approximation and differentials

Definition 2.72 (Tangent line, normal line and linearization). The tangent line to $f$ at $x = a$ is

y = f(a) + f'(a)(x-a),

and the normal line is the line through $(a, f(a))$ perpendicular to the tangent, with slope $-\dfrac{1}{f'(a)}$ when $f'(a) \neq 0$ . The function

L(x) = f(a) + f'(a)(x-a)

is the linearization of $f$ at $a$ , and $f(x) \approx L(x)$ for $x$ near $a$ is the linear approximation.

The linearization is the best straight-line approximation to $f$ near $a$ , in a precise sense: not only is the error $f(x) - L(x)$ small, it is small compared with the distance $x - a$ .

Proposition 2.73 (The error of the linear approximation). If $f$ is differentiable at $a$ , then

\lim_{x\to a}\frac{f(x) - L(x)}{x - a} = 0.

Proof. For $x \neq a$ ,

\frac{f(x) - L(x)}{x-a} = \frac{f(x) - f(a) - f'(a)(x-a)}{x-a} = \frac{f(x)-f(a)}{x-a} - f'(a),

which tends to $f'(a) - f'(a) = 0$ by the definition of the derivative.∎

No other line through $(a, f(a))$ has this property, which is what makes the tangent line special among all lines. The differential is the same idea with the roles renamed.

Definition 2.74 (Differentials). Let $y = f(x)$ be differentiable. The differential $dx$ is an independent variable (any real number), and the differential of $y$ is

dy = f'(x)\,dx.

If $dx = \Delta x$ is a change in $x$ , then $dy$ is the change in height along the tangent line, while $\Delta y = f(x + \Delta x) - f(x)$ is the change along the curve; $dy \approx \Delta y$ for small $\Delta x$ .

This finally gives $dy$ and $dx$ meanings of their own such that their quotient is literally $\frac{dy}{dx} = f'(x)$ , which is what the Leibniz notation promised. In applications $dx$ is a small error or change in a measured quantity and $dy$ the resulting approximate error in a computed one.

Intuition. Near a point, every smooth curve looks almost exactly like its tangent line, so the tangent line is the best straight-line guess for the function a short distance away. Standing on a hillside and taking one step, you predict your change in altitude by the slope where you stand; that prediction is $dy$ , the true change is $\Delta y$ , and the difference is the curvature you ignored — which the second derivative measures.

Example 2.75 (Estimating a square root). Use a linearization to estimate $\sqrt{4.1}$ , and compare with the true value.

Solution.

Take $f(x) = \sqrt{x}$ at $a = 4$ , where $f(4) = 2$ and $f'(x) = \dfrac{1}{2\sqrt{x}}$ gives $f'(4) = \dfrac14$ .
$L(x) = 2 + \tfrac14(x - 4)$ , so $\sqrt{4.1} \approx 2 + \tfrac14(0.1) = 2.025$ .
The true value is $2.02485\ldots$ : an error of about one part in ten thousand. The estimate is slightly too large because $\sqrt{x}$ is concave down, so the tangent lies above the curve.

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Example 2.76 (Tangent and normal lines). Find the tangent and normal lines to $y = x\sqrt{x}$ at $(4, 8)$ .

Solution.

Write $y = x^{3/2}$ , so $y' = \tfrac32 x^{1/2}$ and the slope at $x = 4$ is $\tfrac32\cdot 2 = 3$ .
Tangent: $y - 8 = 3(x - 4)$ , that is $y = 3x - 4$ .
Normal: slope $-\tfrac13$ , so $y - 8 = -\tfrac13(x - 4)$ , that is $y = -\tfrac13 x + \tfrac{28}{3}$ .
Sanity check: both lines pass through $(4, 8)$ , and the product of their slopes is $3\cdot(-\frac13) = -1$ .

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Example 2.77 (Linear approximation of a power). Estimate $(1.98)^5$ without a calculator.

Solution.

Take $f(x) = x^{5}$ at $a = 2$ : $f(2) = 32$ and $f'(x) = 5x^{4}$ gives $f'(2) = 80$ .
With $x - a = -0.02$ : $(1.98)^5 \approx 32 + 80(-0.02) = 32 - 1.6 = 30.4$ .
The true value is $30.43\ldots$ ; the linear approximation is good to about one part in a thousand.

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Example 2.78 (Error propagation with differentials). The radius of a sphere is measured as $21$ cm with a possible error of $0.05$ cm. Estimate the maximum error in the computed volume, and the relative error.

Solution.

$V = \tfrac43\pi r^{3}$ , so $dV = 4\pi r^{2}\,dr$ .
With $r = 21$ and $dr = 0.05$ : $dV = 4\pi(441)(0.05) = 88.2\pi \approx 277$ cm $^3$ .
Relative error: $\dfrac{dV}{V} = \dfrac{4\pi r^{2}\,dr}{\frac43\pi r^{3}} = 3\,\dfrac{dr}{r} = 3\cdot\dfrac{0.05}{21} \approx 0.007$ , about $0.7\%$ .
A relative error in the radius is tripled in the volume, because volume scales as the cube of the radius; the differential makes that factor of $3$ visible.

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Pitfall. $dy$ is the change along the tangent line, $\Delta y$ the change along the curve. They agree only in the limit. For $y = x^{2}$ from $x = 1$ with $dx = 1$ , $dy = 2$ but $\Delta y = 3$ ; the approximation is only useful when $dx$ is small.

Exercises for §2.16 →

2.17Rates of change in the sciences

Definition 2.79 (Average and instantaneous rate of change). If $y = f(x)$ , the average rate of change of $y$ with respect to $x$ over $[x_1, x_2]$ is

\frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1},

and the instantaneous rate of change at $x_1$ is its limit as $x_2 \to x_1$ , namely $f'(x_1)$ .

Every science has quantities that depend on others, and the derivative of one with respect to another is a rate with units: units of $y$ per unit of $x$ . Keeping track of the units is the fastest way to interpret a derivative correctly.

In physics, if $s(t)$ is the position of a particle on a line, then $v(t) = s'(t)$ is its velocity, whose sign gives the direction of motion, $|v(t)|$ is its speed, and $a(t) = v'(t) = s''(t)$ is its acceleration. In chemistry, if $[C](t)$ is the concentration of a product, $\frac{d[C]}{dt}$ is the rate of reaction, and for a first-order decay $[C] = C_0e^{-kt}$ that rate is $-k[C]$ , proportional to what remains. In biology, $\frac{dn}{dt}$ is the growth rate of a population $n(t)$ . In economics, if $C(x)$ is the cost of producing $x$ units then $C'(x)$ is the marginal cost, the approximate cost of producing one more unit:

C'(x) \approx \frac{C(x+1) - C(x)}{1} = C(x+1) - C(x).

Intuition. A derivative always has the units of the top over the units of the bottom, and reading them tells you what it means. If $T(h)$ is temperature in degrees at height $h$ in metres, then $T'(h) = -0.0065$ is $-0.0065$ degrees per metre: climb one metre and it gets $0.0065$ degrees colder. If $C(x)$ is cost in dollars for $x$ items, $C'(500) = 15$ means the 501st item costs about $\$15$ .

Example 2.80 (Motion along a line). A particle moves so that its position at time $t$ seconds is $s(t) = t^{3} - 6t^{2} + 9t$ metres. Find its velocity and acceleration, the times when it is at rest, and the intervals on which it moves in the positive direction.

Solution.

$v(t) = s'(t) = 3t^{2} - 12t + 9 = 3(t-1)(t-3)$ and $a(t) = v'(t) = 6t - 12$ .
At rest when $v = 0$ : $t = 1$ and $t = 3$ .
Moving in the positive direction when $v > 0$ , which by the factored form is $t < 1$ or $t > 3$ . On $(1, 3)$ the particle moves backward.
At $t = 1$ the acceleration is $a(1) = -6$ : the particle is at rest but about to move backward, since its velocity is decreasing through zero. Sanity check: $s(1) = 4$ and $s(3) = 0$ , so it does indeed come back toward the origin between those times.

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Example 2.81 (Growth of a bacterial colony). A colony starts with $100$ bacteria and doubles every hour, so $n(t) = 100\cdot 2^t$ . How fast is it growing after $4$ hours?

Solution.

$n'(t) = 100\cdot 2^t\ln 2$ by the rule for $a^x$ .
At $t = 4$ : $n'(4) = 100\cdot 16\cdot\ln 2 = 1600\ln 2 \approx 1109$ bacteria per hour.
Sanity check: over the fifth hour the colony grows from $1600$ to $3200$ , an average of $1600$ per hour; the instantaneous rate at the start of that hour should be smaller, and $1109$ is.

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Example 2.82 (Marginal cost). A factory's cost of producing $x$ units is $C(x) = 10000 + 5x + 0.01x^{2}$ dollars. Find the marginal cost at $x = 500$ and compare it with the actual cost of the 501st unit.

Solution.

$C'(x) = 5 + 0.02x$ , so $C'(500) = 5 + 10 = 15$ dollars per unit.
The actual cost of the 501st unit is $C(501) - C(500) = 5\cdot 1 + 0.01(501^2 - 500^2) = 5 + 0.01\cdot 1001 = 15.01$ dollars.
The marginal cost approximates the true incremental cost to within one cent, because $C$ is nearly linear over a change of one unit.

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Example 2.83 (A related rate, previewed). Air is pumped into a spherical balloon so that its radius grows at $2$ cm/s. How fast is the volume growing when the radius is $5$ cm?

Solution.

$V = \tfrac43\pi r^{3}$ , and both $V$ and $r$ depend on time. Differentiate with respect to $t$ using the chain rule: $\dfrac{dV}{dt} = 4\pi r^{2}\dfrac{dr}{dt}$ .
With $r = 5$ and $\frac{dr}{dt} = 2$ : $\frac{dV}{dt} = 4\pi\cdot 25\cdot 2 = 200\pi \approx 628$ cm $^3$ /s.
Every related-rates problem has this shape — a geometric relation, differentiated in time by the chain rule — and the applications chapter develops it.

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Pitfall. Velocity and speed are different: velocity $s'(t)$ can be negative, speed $|s'(t)|$ cannot. "The particle is slowing down" means speed is decreasing, which happens when velocity and acceleration have opposite signs, not merely when $a < 0$ .

Exercises for §2.17 →

2.18Interpreting the derivative graphically

The sign of $f'$ reports the direction of travel: $f' > 0$ on an interval means $f$ is increasing there, $f' < 0$ means decreasing, and $|f'|$ says how steep the climb or descent is. (That a positive derivative forces a function to increase is intuitively clear and is proved from the Mean Value Theorem in the applications chapter.)

Definition 2.84 (Critical point). A critical point of $f$ is a number $c$ in the domain of $f$ where $f'(c) = 0$ or $f'(c)$ does not exist.

Where $f'(c) = 0$ the tangent is horizontal. Critical points are the candidates — nothing more — for local maxima and minima. The sign change of $f'$ decides which: if $f'$ goes from $+$ to $-$ at $c$ the function rose then fell, so $c$ is a local maximum; from $-$ to $+$ it is a local minimum. If the sign does not change, it is neither: $f(x) = x^{3}$ flattens at $x = 0$ and keeps climbing.

Reading the graph of $f'$ therefore tells you the shape of $f$ . Where $f'$ lies above the axis, $f$ climbs; where it lies below, $f$ falls; the $x$ -intercepts of $f'$ mark the peaks, valleys and flat spots of $f$ . The reverse translation works too: given $f$ , sketch $f'$ by reading slopes — steep uphill for large positive values, near zero where the graph flattens, steep downhill for large negative ones. The graph of $f'$ is $0$ exactly where $f$ has a horizontal tangent, and $f'$ is undefined at any corner of $f$ .

Intuition. The graph of $f'$ is a steepness report on $f$ : it is the elevation profile of a trail, the gradient you feel underfoot at each point along the walk. Positive means uphill, negative means downhill, zero means a flat moment — a hilltop, a valley, or a brief rest on a slope that continues in the same direction.

Example 2.85 (Increasing and decreasing from the sign of $f'$ ). Find where $f(x) = x^{3} - 3x$ is increasing and where it is decreasing, and classify its critical points.

Solution.

$f'(x) = 3x^{2} - 3 = 3(x-1)(x+1)$ , with zeros at $x = \pm 1$ .
Test the sign on each interval: for $x < -1$ both factors are negative, so $f' > 0$ ; on $(-1, 1)$ the factors have opposite signs, so $f' < 0$ ; for $x > 1$ , $f' > 0$ .
So $f$ is increasing on $(-\infty, -1)$ and $(1, \infty)$ and decreasing on $(-1, 1)$ .
At $x = -1$ the sign changes from $+$ to $-$ : a local maximum, $f(-1) = 2$ . At $x = 1$ it changes from $-$ to $+$ : a local minimum, $f(1) = -2$ .

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Example 2.86 (A critical point that is not an extremum). Classify the critical points of $f(x) = x^{4} - 4x^{3}$ .

Solution.

$f'(x) = 4x^{3} - 12x^{2} = 4x^{2}(x - 3)$ , so the critical points are $x = 0$ and $x = 3$ .
The factor $4x^{2}$ is never negative, so the sign of $f'$ is the sign of $x - 3$ : negative for $x < 3$ (except at $x = 0$ where it is zero) and positive for $x > 3$ .
At $x = 3$ the sign changes from $-$ to $+$ : a local minimum, $f(3) = 81 - 108 = -27$ .
At $x = 0$ the sign is negative on both sides: the graph flattens momentarily and continues downhill. It is a critical point but not an extremum.

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Example 2.87 (Sketching $f'$ from $f$ ). Describe the graph of $f'$ when $f(x) = \sin x$ on $[0, 2\pi]$ , using only the shape of the sine curve.

Solution.

At $x = 0$ the sine curve climbs at its steepest, so $f'$ starts at its maximum value, $1$ .
At $x = \frac{\pi}{2}$ the curve peaks: horizontal tangent, so $f'(\frac{\pi}{2}) = 0$ .
Between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ the curve falls, steepest at $x = \pi$ : so $f'$ is negative, bottoming out at $-1$ when $x = \pi$ .
At $x = \frac{3\pi}{2}$ the curve troughs: $f' = 0$ again, and then $f'$ climbs back to $1$ at $x = 2\pi$ .
The graph of $f'$ described is exactly the cosine curve, in agreement with $\frac{d}{dx}\sin x = \cos x$ .

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Pitfall. $f'(c) = 0$ does not make $c$ a maximum or minimum. It is a candidate. Only a sign change of $f'$ at $c$ (or the second derivative test) settles the question.

Exercises for §2.18 →

2.19Hyperbolic functions and their derivatives

Definition 2.88 (Hyperbolic functions).

\sinh x = \frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}, \qquad \tanh x = \frac{\sinh x}{\cosh x},

with $\operatorname{sech} x = \dfrac{1}{\cosh x}$ , $\operatorname{csch} x = \dfrac{1}{\sinh x}$ and $\coth x = \dfrac{\cosh x}{\sinh x}$ .

They are named for the hyperbola because the point $(\cosh t, \sinh t)$ traces $x^{2} - y^{2} = 1$ , just as $(\cos t, \sin t)$ traces the circle $x^{2} + y^{2} = 1$ : the identity

\cosh^2 x - \sinh^2 x = 1

is verified in one line by expanding $\frac{(e^x+e^{-x})^2 - (e^x - e^{-x})^2}{4} = \frac{4}{4}$ . It is the hyperbolic twin of $\cos^{2} x + \sin^{2} x = 1$ , and the single sign change is what separates the two families. Dividing it by $\cosh^{2} x$ gives $1 - \tanh^{2} x = \operatorname{sech}^2 x$ .

Theorem 2.89 (Derivatives of the hyperbolic functions).

\frac{d}{dx}\sinh x = \cosh x, \qquad \frac{d}{dx}\cosh x = \sinh x, \qquad \frac{d}{dx}\tanh x = \operatorname{sech}^2 x,

\frac{d}{dx}\operatorname{sech} x = -\operatorname{sech} x\tanh x, \qquad \frac{d}{dx}\operatorname{csch} x = -\operatorname{csch} x\coth x, \qquad \frac{d}{dx}\coth x = -\operatorname{csch}^2 x.

Proof. From the definitions and $\frac{d}{dx}e^{-x} = -e^{-x}$ ,

\frac{d}{dx}\sinh x = \frac{e^x - (-e^{-x})}{2} = \frac{e^x + e^{-x}}{2} = \cosh x, \qquad \frac{d}{dx}\cosh x = \frac{e^x + (-e^{-x})}{2} = \sinh x.

For $\tanh x$ the quotient rule gives

\frac{\cosh x\cdot\cosh x - \sinh x\cdot\sinh x}{\cosh^2 x} = \frac{1}{\cosh^2 x} = \operatorname{sech}^2 x,

using the identity. The other three are quotient-rule computations of the same kind.∎

Notice what is missing: the derivative of $\cosh$ has no minus sign. Where the circular functions alternate in sign, $\sinh$ and $\cosh$ simply swap, which is often the fastest way to tell the two families apart. It also means $\sinh$ and $\cosh$ both solve $y'' = y$ , just as $\sin$ and $\cos$ solve $y'' = -y$ .

Because $\sinh$ is increasing on all of $\mathbb{R}$ and $\cosh$ is increasing on $[0,\infty)$ , they have inverses, and since the functions are built from exponentials the inverses can be written with logarithms:

\sinh^{-1}x = \ln\!\left(x + \sqrt{x^2+1}\right), \qquad \cosh^{-1}x = \ln\!\left(x + \sqrt{x^2-1}\right)\ (x \ge 1), \qquad \tanh^{-1}x = \tfrac12\ln\frac{1+x}{1-x}\ (|x|<1).

Theorem 2.90 (Derivatives of the inverse hyperbolic functions).

\frac{d}{dx}\sinh^{-1}x = \frac{1}{\sqrt{x^2+1}}, \qquad \frac{d}{dx}\cosh^{-1}x = \frac{1}{\sqrt{x^2-1}}\ (x > 1), \qquad \frac{d}{dx}\tanh^{-1}x = \frac{1}{1-x^2}\ (|x| < 1).

Proof. Let $y = \sinh^{-1}x$ , so $\sinh y = x$ . Differentiating implicitly, $\cosh y\cdot y' = 1$ , so $y' = \dfrac{1}{\cosh y}$ . Since $\cosh y > 0$ always, $\cosh y = \sqrt{1 + \sinh^{2} y} = \sqrt{1 + x^{2}}$ , giving

\frac{d}{dx}\sinh^{-1}x = \frac{1}{\sqrt{x^2+1}}.

The same result comes from the logarithmic form: with $u = x + \sqrt{x^{2}+1}$ , $u' = 1 + \dfrac{x}{\sqrt{x^{2}+1}} = \dfrac{\sqrt{x^{2}+1} + x}{\sqrt{x^{2}+1}} = \dfrac{u}{\sqrt{x^{2}+1}}$ , so $\dfrac{u'}{u} = \dfrac{1}{\sqrt{x^{2}+1}}$ .

For $\cosh^{-1}$ , the identical argument with $\cosh y = x$ gives $y' = \frac{1}{\sinh y} = \frac{1}{\sqrt{x^{2}-1}}$ , the positive root because $y \ge 0$ . For $\tanh^{-1}$ , $\tanh y = x$ gives $\operatorname{sech}^2 y\cdot y' = 1$ , so $y' = \frac{1}{1 - \tanh^{2} y} = \frac{1}{1-x^{2}}$ .∎

These derivatives are why the hyperbolic functions matter in integration: $\frac{1}{\sqrt{x^{2}+1}}$ and $\frac{1}{\sqrt{x^{2}-1}}$ turn up constantly, and their antiderivatives are inverse hyperbolic functions. Compare with the inverse trigonometric table: $\frac{1}{\sqrt{1-x^{2}}}$ belongs to $\arcsin$ , $\frac{1}{\sqrt{1+x^{2}}}$ to $\sinh^{-1}$ ; the sign under the root is the whole difference.

Intuition. A chain hanging between two posts does not form a parabola but a catenary, $y = a\cosh(x/a)$ . Its shape is determined by the fact that the tension in the chain must balance the weight of the chain below each point, and that balance is the equation $y'' = \frac{1}{a}\sqrt{1 + (y')^2}$ , which $\cosh$ satisfies and no parabola does. Hyperbolic functions are not a curiosity: rapidity in special relativity is a hyperbolic angle, and $\tanh$ is the standard S-shaped switch in neural networks.

Example 2.91 (A hyperbolic composition). Find $\frac{d}{dx}\cosh(3x^{2})$ .

Solution.

Outer function $\cosh u$ with $u = 3x^{2}$ .
Derivative of the outer at the inner: $\sinh(3x^{2})$ . Derivative of the inner: $6x$ .
By the chain rule, $\frac{d}{dx}\cosh(3x^{2}) = 6x\sinh(3x^{2})$ .

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Example 2.92 (Product with hyperbolic functions). Show that $\frac{d}{dx}\left[x\sinh x - \cosh x\right] = x\cosh x$ .

Solution.

Product rule on $x\sinh x$ : $1\cdot\sinh x + x\cosh x$ .
$\frac{d}{dx}(-\cosh x) = -\sinh x$ .
Sum: $\sinh x + x\cosh x - \sinh x = x\cosh x$ . So $x\sinh x - \cosh x$ is an antiderivative of $x\cosh x$ , a fact integration by parts will rediscover.

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Example 2.93 (An inverse hyperbolic composition). Find $\frac{d}{dx}\sinh^{-1}(\tan x)$ for $-\frac{\pi}{2} < x < \frac{\pi}{2}$ , and simplify.

Solution.

Outer $\sinh^{-1}u$ , inner $u = \tan x$ with $u' = \sec^{2} x$ .
$\dfrac{u'}{\sqrt{u^{2}+1}} = \dfrac{\sec^{2} x}{\sqrt{\tan^{2} x + 1}} = \dfrac{\sec^{2} x}{\sqrt{\sec^{2} x}}$ .
On the given interval $\sec x > 0$ , so $\sqrt{\sec^{2} x} = \sec x$ and the derivative is $\sec x$ .
Sanity check at $x = 0$ : $\sinh^{-1}(\tan x) \approx \sinh^{-1}(x) \approx x$ for small $x$ , slope $1$ , and $\sec 0 = 1$ .

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Pitfall. The hyperbolic derivatives have no alternating signs: $\frac{d}{dx}\cosh x = +\sinh x$ . Writing $-\sinh x$ by analogy with $\cos$ is the standard slip. The minus signs in the hyperbolic table belong only to the three reciprocal functions $\operatorname{sech}$ , $\operatorname{csch}$ and $\coth$ .

Exercises for §2.19 →