Derivatives

(a) Combine, factor, cancel. With $a=1$ and $f(1)=1$,

$$f^{\prime }(1)=\underset{x\to 1}{\lim }\frac{\frac{1}{x^2}-1}{x-1}.$$

Put the numerator over a common denominator: $\frac{1}{x^2}-1=\frac{1-x^2}{x^2}=-\frac{(x-1)(x+1)}{x^2}$. For $x\ne 1$ the factor $x-1$ cancels against the denominator, and a limit never inspects $x=1$ itself, so the cancellation is legitimate:

$$\frac{f(x)-f(1)}{x-1}=-\frac{x+1}{x^2}(x\ne 1).$$

The right side is continuous at $1$, so substitute: $f^{\prime }(1)=-\frac{2}{1}=-2$.

Check: the power rule gives $f^{\prime }(x)=-2x^{-3}$, and $f^{\prime }(1)=-2$. The sign is right, since $1/x^2$ is decreasing for $x>0$.

(b) Rationalise. The difference quotient is

$$\frac{\sqrt{2x+2h+3}-\sqrt{2x+3}}{h},$$

of the form $\frac{0}{0}$ as $h\to 0$. Multiply above and below by the conjugate $\sqrt{2x+2h+3}+\sqrt{2x+3}$. The numerator becomes $(2x+2h+3)-(2x+3)=2h$, so for $h\ne 0$ (which is all the limit ever looks at) the quotient equals

$$\frac{2h}{h\left(\sqrt{2x+2h+3}+\sqrt{2x+3}\right)}=\frac{2}{\sqrt{2x+2h+3}+\sqrt{2x+3}}.$$

Let $h\to 0$: the denominator tends to $2\sqrt{2x+3}$, so

$$f^{\prime }(x)=\frac{2}{2\sqrt{2x+3}}=\frac{1}{\sqrt{2x+3}}.$$

Check: the chain rule gives $\frac{1}{2\sqrt{2x+3}}\cdot 2$, the same thing.

Domains. $f$ is defined where $2x+3\ge 0$, that is on $\left[-\frac{3}{2},\infty \right)$. The formula for $f^{\prime }$ needs $2x+3>0$, so the domain of $f^{\prime }$ is $\left(-\frac{3}{2},\infty \right)$. At the endpoint $x=-\frac{3}{2}$ the function is only defined to the right, and even the one-sided quotient fails: $f(-\frac{3}{2}+h)/h=\sqrt{2h}/h=\sqrt{2/h}\to +\infty$. The graph starts there with a vertical tangent, and a vertical line has no slope. This is the general pattern: the domain of $f^{\prime }$ can be strictly smaller than the domain of $f$.

(a) $g(0)=0$. For $h>0$, $|h|=h$ and $g(h)=h^2$, so the quotient is $\frac{h^2-0}{h}=h\to 0$; hence $g_{+}^{\prime }(0)=0$. For $h<0$, $|h|=-h$ and $g(h)=-h^2$, so the quotient is $\frac{-h^2}{h}=-h\to 0$; hence $g_{-}^{\prime }(0)=0$. The one-sided derivatives exist and agree, so $g^{\prime }(0)$ exists and equals $0$. The absolute value did not produce a corner here, because the factor $x$ flattens both branches at the origin.

(b) For $x>0$, $g(x)=x^2$ and $g^{\prime }(x)=2x$. For $x<0$, $g(x)=-x^2$ and $g^{\prime }(x)=-2x$. Together with $g^{\prime }(0)=0$ this is

$$g^{\prime }(x)=2|x|\text{for all }x.$$

Continuity of $g^{\prime }$ at $0$: $2|x|\to 0=g^{\prime }(0)$ as $x\to 0$, so yes.

Differentiability of $g^{\prime }$ at $0$: the difference quotient of $g^{\prime }$ is

$$\frac{g^{\prime }(h)-g^{\prime }(0)}{h}=\frac{2|h|}{h}=\{\begin{array}{ll}2,& h>0,\\ -2,& h<0,\end{array}$$

so the one-sided derivatives of $g^{\prime }$ at $0$ are $2$ and $-2$. They differ, so $g^{\prime \prime }(0)$ does not exist: the graph of $g^{\prime }=2|x|$ has a corner at the origin. Summary: $g$ is differentiable everywhere, but it is not twice differentiable at $0$. Being differentiable once does not make a function smooth.

(c) Continuity: $|p(x)-p(0)|=\sqrt{|x|}\to 0$ as $x\to 0$, so $p$ is continuous at $0$.

The quotients: $p(0)=0$, so the difference quotient is $\frac{\sqrt{|h|}}{h}$. For $h>0$ this is $\frac{\sqrt{h}}{h}=\frac{1}{\sqrt{h}}\to +\infty$. For $h<0$, write $h=-|h|$: the quotient is $\frac{\sqrt{|h|}}{-|h|}=-\frac{1}{\sqrt{|h|}}\to -\infty$.

The two one-sided quotients tend to infinity with opposite signs. That is the signature of a cusp, on the model of $x^{2/3}$: the graph comes to a sharp point at the origin with a vertical tangent direction on each side. It is not a corner (those have finite, unequal one-sided slopes) and not a vertical tangent (those have infinite one-sided quotients of the same sign, like $\sqrt[3]{x}$).

(a) Simplify first. Divide each term by $x^2$ and write everything as a power:

$$y=2x-5x^{-1}+4x^{1/2-2}=2x-5x^{-1}+4x^{-3/2}.$$

Now the sum and power rules apply term by term:

$$y^{\prime }=2+5x^{-2}+4\cdot \left(-\frac{3}{2}\right)x^{-5/2}=2+\frac{5}{x^2}-\frac{6}{x^2\sqrt{x}}.$$

Sign check on the middle term: $-5x^{-1}=-5/x$ is increasing for $x>0$, so its derivative $+5/x^2$ should be positive, and it is.

(b) Horizontal tangent means $y^{\prime }=0$. By the power rule $y^{\prime }=4x^3-4x=4x(x^2-1)=4x(x-1)(x+1)$, which vanishes at $x=0$, $x=1$ and $x=-1$. The corresponding heights are $y(0)=0$ and $y(\pm 1)=1-2=-1$, so the points are

$$(0,0),(1,-1),(-1,-1).$$

(The curve is even, so the two outer points had to be mirror images.)

(c) Parametrise by the point of tangency. The tangent at $x=a$ touches the parabola at $(a,a^2+1)$ and has slope $2a$, so its equation is

$$y=a^2+1+2a(x-a).$$

Demand that $(1,-2)$ lies on it: $-2=a^2+1+2a(1-a)=-a^2+2a+1$, that is $a^2-2a-3=0$, so $(a-3)(a+1)=0$ and $a=3$ or $a=-1$.

For $a=3$: the point is $(3,10)$, the slope is $6$, and the line is $y=10+6(x-3)=6x-8$.

For $a=-1$: the point is $(-1,2)$, the slope is $-2$, and the line is $y=2-2(x+1)=-2x$.

Check both through $(1,-2)$: $6\cdot 1-8=-2$ and $-2\cdot 1=-2$. The common mistake is to differentiate at $x=1$ and use slope $2$; but $(1,-2)$ is not on the curve, so nothing is tangent there. The unknown is the point of contact, not the given point.

(a) The slope at $x=a$ is $e^a$, so the tangent line is

$$y=e^a+e^a(x-a)=e^a(x-a+1).$$

Set $y=0$. Since $e^a>0$ for every $a$ we may divide by it, leaving $x-a+1=0$, that is $x=a-1$. $■$

Geometrically: from any point of the exponential curve, run the tangent down to the axis and you always land exactly one unit to the left of the foot of the point. That fixed horizontal run of $1$ is another way of saying "slope equals height": rise $e^a$ over run $1$ is slope $e^a$.

(b) A tangent through $(2,0)$ is a tangent whose $x$-intercept is $2$, so by (a) $a-1=2$ and $a=3$. It touches the curve at $(3,e^3)$ and is

$$y=e^3(x-2).$$

Check: at $x=3$ the line gives $e^3$, which is on the curve, and its slope $e^3$ is the derivative there. The tangent through the origin is the special case $a=1$, giving $y=ex$.

(c) Put $x=0$ in the tangent line: the $y$-intercept is

$$e^a(1-a).$$

As $e^a>0$, the sign is the sign of $1-a$: the intercept is negative for $a>1$, zero for $a=1$ (the tangent through the origin again), and positive for $a<1$. In words: tangents drawn to the right of $x=1$ dive below the origin before reaching the $y$-axis, and tangents drawn to the left cross the $y$-axis above it. Every one of them has positive slope $e^a$, so each is an increasing line; only where it sits changes.

(a) Group the product as $(fg)\cdot h$. By the product rule, $fg$ is differentiable at $x$ (its hypotheses, that $f$ and $g$ are differentiable at $x$, are given) with $(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$. Now apply the product rule again to the two differentiable factors $fg$ and $h$:

$$(fgh{)}^{\prime }=((fg)h{)}^{\prime }=(fg{)}^{\prime }h+(fg)h^{\prime }=(f^{\prime }g+f{g}^{\prime })h+fg{h}^{\prime }=f^{\prime }gh+f{g}^{\prime }h+fg{h}^{\prime }.■$$

The shape is worth noticing: one term for each factor, that factor differentiated and the others left alone. The same argument by induction gives the rule for any number of factors.

(b) Take $f=x$, $g=e^x$, $h=\sin x$, with $f^{\prime }=1$, $g^{\prime }=e^x$, $h^{\prime }=\cos x$. By (a),

$$y^{\prime }=1\cdot e^x\sin x+x\cdot e^x\sin x+x{e}^x\cos x=e^x[(1+x)\sin x+x\cos x].$$

At $x=0$: $e^0[1\cdot 0+0]=0$. At $x=\pi$: $\sin \pi =0$ and $\cos \pi =-1$, so $y^{\prime }(\pi )=e^{\pi }[0+\pi \cdot (-1)]=-\pi e^{\pi }$.

Sanity check on $y^{\prime }(0)=0$: near $0$, $e^x\approx 1$ and $\sin x\approx x$, so $y\approx x^2$, which is flat at the origin.

(c) Proof. Since $f(a)=0$, the product has $(fg)(a)=0$, and for $h\ne 0$

$$\frac{f(a+h)g(a+h)-f(a)g(a)}{h}=\frac{f(a+h)}{h}g(a+h)=\frac{f(a+h)-f(a)}{h}\cdot g(a+h),$$

where the last step inserts $-f(a)=0$ into the numerator. As $h\to 0$ the first factor tends to $f^{\prime }(a)$, by the definition of the derivative, and the second tends to $g(a)$, by continuity of $g$ at $a$. Both limits exist as real numbers, so the product law for limits applies and the difference quotient tends to $f^{\prime }(a)g(a)$. That is exactly the statement that $fg$ is differentiable at $a$ with $(fg{)}^{\prime }(a)=f^{\prime }(a)g(a)$. $■$

Application. Take $f(x)=\sin x$ and $g(x)=|x|$ at $a=0$: $f(0)=0$, $f^{\prime }(0)=\cos 0=1$, and $|x|$ is continuous at $0$ with $g(0)=0$. Hence $F$ is differentiable at $0$ and $F^{\prime }(0)=1\cdot 0=0$.

The product rule could not be invoked because one of its hypotheses fails: $|x|$ is not differentiable at $0$ (it has a corner there). The rule is silent, not wrong; the definition still answers. Away from $0$ both factors are differentiable and the product rule gives $F^{\prime }(x)=\cos x|x|+\sin x\cdot \mathrm{sgn}(x)$, which also tends to $0$ as $x\to 0$ from either side, consistent with what we found.

(a) Write $\csc x=\frac{1}{\sin x}$ and apply the quotient rule with $f=1$, $g=\sin x$, $f^{\prime }=0$, $g^{\prime }=\cos x$, valid wherever $\sin x\ne 0$:

$$\frac{d}{dx}\frac{1}{\sin x}=\frac{0\cdot \sin x-1\cdot \cos x}{{\sin }^{2}x}=-\frac{\cos x}{{\sin }^{2}x}=-\frac{1}{\sin x}\cdot \frac{\cos x}{\sin x}=-\csc x\cot x.■$$

The minus sign is the co-function rule at work: $\csc$ is the co-function of $\sec$, whose derivative is $+\sec x\tan x$.

(b) Differentiate term by term, using the product rule on the first two.

$\frac{d}{dx}\left(x^2\sin x\right)=2x\sin x+x^2\cos x$.

$\frac{d}{dx}\left(2x\cos x\right)=2\cos x-2x\sin x$.

$\frac{d}{dx}(-2\sin x)=-2\cos x$.

Add:

$$y^{\prime }=\underline{2x\sin x}+x^2\cos x+\underline{\underline{2\cos x}}-\underline{2x\sin x}-\underline{\underline{2\cos x}}=x^2\cos x.$$

Two pairs cancel and a single term survives. (This is the kind of cancellation that integration by parts produces: $y$ is an antiderivative of $x^2\cos x$.)

(c) Product rule: $y^{\prime }=\cos x\cdot \cos x+\sin x\cdot (-\sin x)={\cos }^{2}x-{\sin }^{2}x$.

A horizontal tangent needs ${\cos }^{2}x={\sin }^{2}x$. If $\cos x=0$ then ${\sin }^{2}x=1\ne 0$, so no solution has $\cos x=0$, and we may divide: ${\tan }^{2}x=1$, that is $\tan x=\pm 1$. In $[0,2\pi ]$ this gives

$$x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}.$$

Values: at $\frac{\pi }{4}$ and $\frac{5\pi }{4}$ the sine and cosine have the same sign and each has magnitude $\frac{\sqrt{2}}{2}$, so $y=\frac{1}{2}$; at $\frac{3\pi }{4}$ and $\frac{7\pi }{4}$ they have opposite signs, so $y=-\frac{1}{2}$.

Cross-check: ${\cos }^{2}x-{\sin }^{2}x=\cos 2x$, and $y=\frac{1}{2}\sin 2x$ has derivative $\cos 2x$ by the chain rule; it vanishes when $2x$ is an odd multiple of $\frac{\pi }{2}$, which is the same list. The maxima $\frac{1}{2}$ and minima $-\frac{1}{2}$ are the peaks and troughs of $\frac{1}{2}\sin 2x$.

(a) Wrong: a missing innermost derivative. The student peeled only two layers. There are three: outer $u^3$, middle $\cos v$, inner $v=x^2$. The derivative of the middle layer, $-\sin (x^2)$, must itself be multiplied by the derivative of what is inside it, $\frac{d}{dx}{x}^2=2x$. The chain rule says the derivative of the inner function is $\frac{d}{dx}\cos (x^2)=-\sin (x^2)\cdot 2x$, not $-\sin (x^2)$. Correctly,

$$y^{\prime }=3{\cos }^2(x^2)\cdot (-\sin (x^2))\cdot 2x=-6x{\cos }^2(x^2)\sin (x^2).$$

A check that exposes the error without any computation: $y$ is an even function of $x$, and the derivative of an even function is odd. The student's answer $-3{\cos }^2(x^2)\sin (x^2)$ is even, so it cannot be the derivative of $y$; the corrected answer carries the odd factor $x$.

(b) Wrong: the outer derivative is evaluated at the wrong point. The chain rule reads $(f\circ g{)}^{\prime }(1)=f^{\prime }(g(1))\cdot g^{\prime }(1)$, and $g(1)=3$, so the outer derivative that matters is $f^{\prime }(3)$, not $f^{\prime }(1)$. The number $f^{\prime }(1)=5$ is the slope of $f$ at an input $f$ never receives here. Correctly,

$$(f\circ g{)}^{\prime }(1)=f^{\prime }(3)g^{\prime }(1)=(-4)(2)=-8.$$

(c) Correct. Rewriting the reciprocal as a negative power is legal, and the power case of the chain rule, $\frac{d}{dx}[g(x){]}^n=n[g(x){]}^{n-1}{g}^{\prime }(x)$ with $g(x)=x^3+1$, $n=-2$ and $g^{\prime }(x)=3x^2$, gives exactly $-2(x^3+1{)}^{-3}\cdot 3x^2$. The quotient rule was not required.

Second method, by the quotient rule with $f=1$ and $g=(x^3+1{)}^2$, $g^{\prime }=2(x^3+1)\cdot 3x^2$:

$$\frac{0\cdot g-1\cdot g^{\prime }}{g^2}=-\frac{6x^2(x^3+1)}{(x^3+1{)}^4}=-\frac{6x^2}{(x^3+1{)}^3},$$

the same result. The lesson of the three parts together: the chain rule fails in exactly two ways — a layer forgotten, or a derivative evaluated at the wrong input — and a correct answer that looks unfamiliar is still correct.

(a) Differentiate each term with respect to $x$, treating $y$ as a function of $x$.

$\frac{d}{dx}{x}^2=2x$. For $xy$ the product rule gives $1\cdot y+x\cdot y^{\prime }$. For $y^2$ the chain rule (outer $u^2$, inner $u=y$) gives $2y{y}^{\prime }$. The right side is a constant, with derivative $0$. So

$$2x-(y+x{y}^{\prime })+2y{y}^{\prime }=0.$$

Collect the $y^{\prime }$ terms: $y^{\prime }(2y-x)=y-2x$, hence

$$\frac{dy}{dx}=\frac{y-2x}{2y-x}(2y\ne x).$$

(b) At $(-1,1)$: $1-(-1)(1)+1=3$, so the point is on $C$. The slope is

$$\frac{1-2(-1)}{2(1)-(-1)}=\frac{3}{3}=1.$$

Tangent: $y-1=1\cdot (x+1)$, that is $y=x+2$. Normal: slope $-1$, so $y-1=-(x+1)$, that is $y=-x$.

Sanity check: the normal passes through the origin, the centre of the ellipse. That is right, because $(-1,1)$ lies on the line $y=-x$, which is an axis of symmetry of this ellipse (the equation is unchanged when $x$ and $y$ are swapped, and also when both are negated), and at the end of an axis the normal points straight at the centre.

(c) Horizontal: the numerator vanishes, $y=2x$. Substitute into $C$: $x^2-2x^2+4x^2=3x^2=3$, so $x=\pm 1$ and the points are $(1,2)$ and $(-1,-2)$. There the denominator is $2y-x=3x=\pm 3\ne 0$, so the slope is genuinely $0$.

Vertical: the denominator vanishes, $x=2y$. Substitute: $4y^2-2y^2+y^2=3y^2=3$, so $y=\pm 1$ and the points are $(2,1)$ and $(-2,-1)$. There the numerator is $y-2x=y-4y=-3y=\mp 3\ne 0$, so the fraction really blows up and the tangent is vertical rather than undetermined.

The four points are the top, bottom, right and left extremes of the ellipse: $y$ ranges over $[-2,2]$ and $x$ over $[-2,2]$.

(d) Differentiate $y^{\prime }=\frac{y-2x}{2y-x}$ by the quotient rule, remembering that $y$ and hence $y^{\prime }$ depend on $x$:

$$y^{\prime \prime }=\frac{(y^{\prime }-2)(2y-x)-(y-2x)(2y^{\prime }-1)}{(2y-x{)}^2}.$$

At $(1,2)$ we know from (c) that $y^{\prime }=0$ and $y-2x=0$, while $2y-x=3$. The second product in the numerator is therefore $0$, and

$$y^{\prime \prime }{|}_{(1,2)}=\frac{(0-2)(3)-0}{3^2}=-\frac{6}{9}=-\frac{2}{3}.$$

It is negative, so the curve is concave down at $(1,2)$: the tangent is horizontal and the curve bends away beneath it. That is what the top of a closed oval must do, and it confirms that $(1,2)$ is the highest point of $C$ rather than the lowest.

(a) $f^{\prime }(x)=5x^4+6x^2+1\ge 1>0$ for every $x$, so $f$ is strictly increasing on $\mathbb{R}$, hence one-to-one, and being a polynomial it is continuous. So $f^{-1}$ exists, and the inverse function theorem applies at every point because $f^{\prime }$ is never $0$.

To use the theorem at $4$ we need $f^{-1}(4)$, that is the solution of $x^5+2x^3+x=4$. By inspection $x=1$ works ($1+2+1=4$), and there is no other solution because $f$ is one-to-one. Then $f^{\prime }(1)=5+6+1=12$ and

$${\left(f^{-1}\right)}^{\prime }(4)=\frac{1}{f^{\prime }(f^{-1}(4))}=\frac{1}{f^{\prime }(1)}=\frac{1}{12}.$$

The number $\frac{1}{f^{\prime }(4)}=\frac{1}{5\cdot 256+6\cdot 16+1}=\frac{1}{1377}$ is the reciprocal of the slope of $f$ at the wrong point. Reflection in $y=x$ sends the point $(1,4)$ on the graph of $f$ to the point $(4,1)$ on the graph of $f^{-1}$, so the slope of $f^{-1}$ at input $4$ is tied to the slope of $f$ at input $1$, not at input $4$.

(b) Oddness says $F(-x)=-F(x)$ for all $x$. Both sides are differentiable functions of $x$, so differentiate both. On the left, the chain rule with inner function $-x$ gives $F^{\prime }(-x)\cdot (-1)$; on the right, $-F^{\prime }(x)$. So $-F^{\prime }(-x)=-F^{\prime }(x)$, that is $F^{\prime }(-x)=F^{\prime }(x)$ for all $x$: $F^{\prime }$ is even. $■$

(Check on our $f$: it is odd, having only odd powers, and $f^{\prime }$ has only even powers.)

(c) $f^{-1}$ is odd. Let $x$ be any real number and put $y=f^{-1}(x)$, so $f(y)=x$. Since $f$ is odd, $f(-y)=-f(y)=-x$. Applying $f^{-1}$ to both sides, $f^{-1}(-x)=-y=-f^{-1}(x)$. $■$

${\left(f^{-1}\right)}^{\prime }$ is even. By the inverse function theorem ($f$ continuous, one-to-one on an interval, $f^{\prime }\ne 0$ everywhere) $f^{-1}$ is differentiable on all of $\mathbb{R}$. It is odd by the first half of (c), so by (b) applied to $F=f^{-1}$ its derivative is even. $■$

One can also see it directly from the formula: ${\left(f^{-1}\right)}^{\prime }(-x)=\frac{1}{f^{\prime }(f^{-1}(-x))}=\frac{1}{f^{\prime }(-f^{-1}(x))}=\frac{1}{f^{\prime }(f^{-1}(x))}={\left(f^{-1}\right)}^{\prime }(x)$, using oddness of $f^{-1}$ and then evenness of $f^{\prime }$.

Consequence. ${\left(f^{-1}\right)}^{\prime }(-4)={\left(f^{-1}\right)}^{\prime }(4)=\frac{1}{12}$. Check by hand: $f(-1)=-4$ and $f^{\prime }(-1)=12$, so the theorem gives $\frac{1}{12}$ directly, as it must.

(a) On $(0,\pi )$ we have $\sin x>0$, so $y>0$ and $\ln y=x\ln (\sin x)$. Differentiate implicitly. The left side becomes $\frac{y^{\prime }}{y}$ by the chain rule. The right side is a product, and $\frac{d}{dx}\ln (\sin x)=\frac{\cos x}{\sin x}=\cot x$ by the chain rule, so

$$\frac{y^{\prime }}{y}=\ln (\sin x)+x\cot x,y^{\prime }=(\sin x{)}^x[\ln (\sin x)+x\cot x].$$

At $x=\frac{\pi }{2}$: $\sin x=1$, so $\ln (\sin x)=0$, and $\cot \frac{\pi }{2}=0$; hence $y^{\prime }\left(\frac{\pi }{2}\right)=1\cdot [0+0]=0$. The value makes sense: for $0<x<\pi$ the base satisfies $0<\sin x\le 1$ and the exponent is positive, so $y\le 1$, with equality exactly at $x=\frac{\pi }{2}$. That is the maximum of $y$, and a maximum in the interior of the domain must have a horizontal tangent.

(b) For $x>0$, $y=x^{\ln x}>0$ and $\ln y=\ln x\cdot \ln x=(\ln x{)}^2$. Differentiating, with the chain rule on both sides,

$$\frac{y^{\prime }}{y}=2\ln x\cdot \frac{1}{x},y^{\prime }=\frac{2\ln x}{x}{x}^{\ln x}.$$

Since $x^{\ln x}>0$ and $x>0$, the sign of $y^{\prime }$ is the sign of $\ln x$: negative on $(0,1)$, zero at $x=1$, positive on $(1,\infty )$. The only horizontal tangent is at $(1,1)$, and because $y^{\prime }$ changes from $-$ to $+$ there, it is a minimum.

Cross-check: $x^{\ln x}=e^{(\ln x{)}^2}\ge e^0=1$, with equality only when $\ln x=0$. The chain rule on this form gives $e^{(\ln x{)}^2}\cdot 2\ln x\cdot \frac{1}{x}$, the same derivative.

(c) Since $f$ and $g$ are nonzero, $|y|=|f|/|g|>0$ and

$$\ln |y|=\ln |f|-\ln |g|.$$

Differentiate, using $\frac{d}{dx}\ln |u|=\frac{u^{\prime }}{u}$ for each of $y$, $f$, $g$:

$$\frac{y^{\prime }}{y}=\frac{f^{\prime }}{f}-\frac{g^{\prime }}{g}.$$

Multiply through by $y=\frac{f}{g}$:

$$y^{\prime }=\frac{f}{g}\left(\frac{f^{\prime }}{f}-\frac{g^{\prime }}{g}\right)=\frac{f^{\prime }}{g}-\frac{f{g}^{\prime }}{g^2}=\frac{f^{\prime }g-f{g}^{\prime }}{g^2}.■$$

What was assumed. Two things the quotient rule does not require. First, $f\ne 0$: the quotient rule needs only $g\ne 0$, but taking $\ln |f|$ needs $f\ne 0$ as well. Second, the differentiability of $y$ itself: we wrote $\frac{d}{dx}\ln |y|=\frac{y^{\prime }}{y}$, which presupposes $y^{\prime }$ exists, whereas the quotient rule proves that $f/g$ is differentiable. So this derivation is a way of remembering or recovering the formula, not an independent proof of the theorem. Its real content is the bookkeeping identity: the relative rate of change of a quotient is the difference of the relative rates of change of numerator and denominator.

(a) By the product rule, $f^{\prime }(x)=1\cdot e^x+x{e}^x=(x+1)e^x$. Again, $f^{\prime \prime }(x)=1\cdot e^x+(x+1)e^x=(x+2)e^x$, and $f^{\prime \prime \prime }(x)=e^x+(x+2)e^x=(x+3)e^x$. Conjecture:

$$f^{(n)}(x)=(x+n)e^x(n\ge 1).$$

Base case $n=1$. $f^{\prime }(x)=(x+1)e^x$, computed above.

Inductive step. Suppose $f^{(n)}(x)=(x+n)e^x$ for some $n\ge 1$. Then, by definition of the $(n+1)$st derivative and the product rule,

$$f^{(n+1)}(x)=\frac{d}{dx}[(x+n)e^x]=1\cdot e^x+(x+n)e^x=(x+(n+1))e^x,$$

which is the formula with $n+1$ in place of $n$. By induction it holds for all $n\ge 1$. $■$

Hence $f^{(10)}(0)=(0+10)e^0=10$.

(b) Base case $n=1$. $\frac{d}{dx}\sin x=\cos x$, and $\cos x=\sin \left(x+\frac{\pi }{2}\right)$ (shifting the sine a quarter period to the left gives the cosine; equivalently, the addition formula gives $\sin x\cos \frac{\pi }{2}+\cos x\sin \frac{\pi }{2}=\cos x$). So the statement holds for $n=1$.

Inductive step. Suppose $\frac{d^n}{d{x}^n}\sin x=\sin \left(x+\frac{n\pi }{2}\right)$. Differentiate once more, using the chain rule with inner function $x+\frac{n\pi }{2}$ (whose derivative is $1$) and then the base-case identity:

$$\frac{d^{n+1}}{d{x}^{n+1}}\sin x=\cos \left(x+\frac{n\pi }{2}\right)=\sin \left(x+\frac{n\pi }{2}+\frac{\pi }{2}\right)=\sin \left(x+\frac{(n+1)\pi }{2}\right).$$

By induction the formula holds for all $n\ge 1$. $■$

With $n=50$: $\sin \left(x+25\pi \right)=\sin \left(x+\pi +24\pi \right)=\sin (x+\pi )=-\sin x$, since $24\pi$ is a whole number of periods. Check against the four-step cycle $\sin \to \cos \to -\sin \to -\cos \to \sin$: $50=48+2$, so the 50th derivative is the second, namely $-\sin x$.

(c) Product rule with the chain rule on $e^{2x}$ (derivative $2e^{2x}$):

$$y^{\prime }=B{e}^{2x}+(A+Bx)\cdot 2e^{2x}=(2A+B+2Bx)e^{2x}.$$

Again:

$$y^{\prime \prime }=2B{e}^{2x}+(2A+B+2Bx)\cdot 2e^{2x}=(4A+4B+4Bx)e^{2x}.$$

Now substitute and factor out $e^{2x}$:

$$y^{\prime \prime }-4y^{\prime }+4y=e^{2x}[(4A+4B+4Bx)-4(2A+B+2Bx)+4(A+Bx)].$$

Collect the bracket by type of term. Constants: $4A+4B-8A-4B+4A=0$. Terms in $x$: $4Bx-8Bx+4Bx=0$. The bracket is identically zero, so $y^{\prime \prime }-4y^{\prime }+4y=0$ for every $A$ and $B$. $■$

Both $e^{2x}$ (take $B=0$) and $x{e}^{2x}$ (take $A=0$) solve the equation on their own; the two-parameter family is their general combination. The factor $x$ appears because $2$ is a repeated root of $r^2-4r+4=(r-2{)}^2$, a fact the differential equations chapter will explain.

(a) $|x-2|=x-2$ for $x\ge 2$ and $2-x$ for $x<2$, so

$$f(x)=\{\begin{array}{ll}2x-x^2,& x<2,\\ x^2-2x,& x\ge 2.\end{array}$$

On the open pieces the power rule applies: $f^{\prime }(x)=2-2x$ for $x<2$ and $f^{\prime }(x)=2x-2$ for $x>2$.

At $x=2$ both formulas give $f(2)=0$, so $f$ is continuous there, and the one-sided derivatives come from the one-sided formulas: $f_{-}^{\prime }(2)=2-2\cdot 2=-2$ and $f_{+}^{\prime }(2)=2\cdot 2-2=2$. (Formally, for $h<0$ the quotient $\frac{f(2+h)-0}{h}=\frac{(2+h)(-h)}{h}=-(2+h)\to -2$, and for $h>0$ it is $\frac{(2+h)h}{h}=2+h\to 2$.) They differ, so $f^{\prime }(2)$ does not exist and $f$ has a corner at $x=2$. Since $2$ is in the domain of $f$ and $f^{\prime }(2)$ does not exist, $x=2$ is a critical point.

$x=0$ is not a corner. The absolute value $|x-2|$ is smooth except at $x=2$; near $x=0$ we are inside the single piece $2x-x^2$, so $f^{\prime }(0)=2$ exists. The zero of $f$ at $0$ is an ordinary crossing with slope $2$.

(b) Critical points are where $f^{\prime }=0$ or $f^{\prime }$ fails to exist. $2-2x=0$ gives $x=1$, which lies in the piece $x<2$, so it counts; $2x-2=0$ gives $x=1$ again, which is not in the piece $x>2$, so the right piece contributes no zero. Together with $x=2$ from (a), the critical points are $x=1$ and $x=2$.

Sign of $f^{\prime }$: on $(-\infty ,1)$, $2-2x>0$; on $(1,2)$, $2-2x<0$; on $(2,\infty )$, $2x-2>0$.