Techniques of Differentiation — Worked Solutions

Ques­tion 1: The chain rule on four func­tions#

The prob­lem#

Dif­fer­en­ti­ate with respect to $x$:

(a) $\cos(5x - 2)$ (b) $(3x^2 - x + 1)^4$ (c) $\sin^3 x$ (d) $\sqrt{1 + e^{2x}}$

Under­stand­ing the prob­lem#

Each func­tion is a com­po­si­tion: one func­tion applied to another. You need $\displaystyle \frac{dy}{dx}$ for each.

The idea#

Use the chain rule: if $y = f(u)$ with $u = g(x)$, then $\displaystyle \frac{dy}{dx} = f'(u)\cdot\frac{du}{dx}$. In each part, first name the inside $u$.

Step-by-step solu­tion#

Part (a)

Step 1. Inside: $u = 5x - 2$, so $\displaystyle \frac{du}{dx} = 5$. Out­side: $\cos u$, whose deriv­a­tive is $-\sin u$.

Step 2. Mul­ti­ply.

$$\displaystyle \frac{d}{dx}\cos(5x - 2) = -\sin(5x - 2)\cdot 5 = -5\sin(5x - 2)$$

Part (b)

Step 1. Inside: $u = 3x^2 - x + 1$, so $\displaystyle \frac{du}{dx} = 6x - 1$. Out­side: $u^4$, deriv­a­tive $4u^3$.

Step 2. Mul­ti­ply.

$$\displaystyle \frac{d}{dx}(3x^2 - x + 1)^4 = 4(3x^2 - x + 1)^3(6x - 1)$$

Part (c)

Step 1. $\sin^3 x$ means $(\sin x)^3$. Inside: $u = \sin x$, $\displaystyle \frac{du}{dx} = \cos x$. Out­side: $u^3$, deriv­a­tive $3u^2$.

Step 2. Mul­ti­ply.

$$\displaystyle \frac{d}{dx}\sin^3 x = 3\sin^2 x\cos x$$

Part (d)

Step 1. Here there are two lay­ers. Out­er­most: $\sqrt{v}$ with $v = 1 + e^{2x}$; its deriv­a­tive is $\displaystyle \frac{1}{2\sqrt{v}}$.

Step 2. Deriv­a­tive of the inside $v = 1 + e^{2x}$: the $1$ gives $0$, and $e^{2x}$ needs the chain rule again, giv­ing $e^{2x}\cdot 2$.

$$\displaystyle \frac{dv}{dx} = 2e^{2x}$$

Step 3. Mul­ti­ply and sim­plify; the $2$s can­cel.

$$\displaystyle \frac{d}{dx}\sqrt{1 + e^{2x}} = \frac{1}{2\sqrt{1 + e^{2x}}}\cdot 2e^{2x} = \frac{e^{2x}}{\sqrt{1 + e^{2x}}}$$

Check­ing the answer#

For (a), at $\displaystyle x = \tfrac{2}{5}$ the inside is $0$ and $\sin 0 = 0$, so the slope is $0$; $\cos$ has a peak there, as expected. For (c), at $x = 0$ the deriv­a­tive is $0$, which fits because $\sin^3 x$ is very flat near $0$.

Answer#

(a) $-5\sin(5x - 2)$; (b) $4(3x^2 - x + 1)^3(6x - 1)$; (c) $3\sin^2 x\cos x$; (d) $\displaystyle \frac{e^{2x}}{\sqrt{1 + e^{2x}}}$.

Com­mon mis­take to avoid#

For­get­ting the "deriv­a­tive of the inside", e.g. writ­ing $-\sin(5x - 2)$ in (a) with­out the fac­tor $5$.

Ques­tion 2: Expo­nen­tial and log­a­rith­mic func­tions, prod­ucts and quo­tients#

The prob­lem#

Dif­fer­en­ti­ate: (a) $e^{x^2}$ (b) $\log(\sin x)$ (c) $x^2 e^x$ (d) $\displaystyle \frac{\log x}{x}$

Under­stand­ing the prob­lem#

(a) and (b) are com­po­si­tions (chain rule). (c) is a prod­uct of two func­tions, and (d) is a quo­tient. Here $\log$ means the nat­ural log­a­rithm, with $\displaystyle \frac{d}{dx}\log x = \frac{1}{x}$.

The idea#

Use the stan­dard deriv­a­tives $\displaystyle \frac{d}{dx}e^x = e^x$ and $\displaystyle \frac{d}{dx}\log x = \frac{1}{x}$, together with the chain rule, the prod­uct rule $(uv)' = u'v + uv'$ and the quo­tient rule $\displaystyle \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$.

Step-by-step solu­tion#

Part (a)

Step 1. Inside $u = x^2$, $\displaystyle \frac{du}{dx} = 2x$. Out­side $e^u$, deriv­a­tive $e^u$.

$$\displaystyle \frac{d}{dx}e^{x^2} = e^{x^2}\cdot 2x = 2xe^{x^2}$$

Part (b)

Step 1. Inside $u = \sin x$, $\displaystyle \frac{du}{dx} = \cos x$. Out­side $\log u$, deriv­a­tive $\displaystyle \frac{1}{u}$.

Step 2. Mul­ti­ply and recog­nise $\displaystyle \frac{\cos x}{\sin x}$.

$$\displaystyle \frac{d}{dx}\log(\sin x) = \frac{1}{\sin x}\cdot\cos x = \cot x$$

Part (c)

Step 1. Prod­uct with $u = x^2$ ($u' = 2x$) and $v = e^x$ ($v' = e^x$).

Step 2. Apply the prod­uct rule and take out the com­mon fac­tor $e^x$.

$$\displaystyle \frac{d}{dx}(x^2e^x) = 2x\cdot e^x + x^2\cdot e^x = e^x(x^2 + 2x)$$

Part (d)

Step 1. Quo­tient with $u = \log x$ ($\displaystyle u' = \tfrac{1}{x}$) and $v = x$ ($v' = 1$).

Step 2. Apply the quo­tient rule.

$$\displaystyle \frac{d}{dx}\frac{\log x}{x} = \frac{\tfrac{1}{x}\cdot x - \log x\cdot 1}{x^2} = \frac{1 - \log x}{x^2}$$

Check­ing the answer#

In (d), the deriv­a­tive is $0$ when $\log x = 1$, i.e. $x = e$. Indeed $\displaystyle \frac{\log x}{x}$ has its largest value at $x = e$, so a zero slope there makes sense.

Answer#

(a) $2xe^{x^2}$; (b) $\cot x$; (c) $e^x(x^2 + 2x)$; (d) $\displaystyle \frac{1 - \log x}{x^2}$.

Ques­tion 3: Inverse trigono­met­ric func­tions#

The prob­lem#

Dif­fer­en­ti­ate: (a) $\sin^{-1}(2x)$ (b) $\tan^{-1}(e^x)$ (c) $\displaystyle \cos^{-1}\frac{1 - x^2}{1 + x^2}$ for $0 < x < 1$.

Under­stand­ing the prob­lem#

Parts (a) and (b) use the stan­dard inverse-trig deriv­a­tives with the chain rule. Part (c) looks heavy, but, as in Exam­ple 3, a sub­sti­tu­tion sim­pli­fies it before dif­fer­en­ti­at­ing. The con­di­tion $0 < x < 1$ tells you which branch to use.

The idea#

Recall $\displaystyle \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$ and $\displaystyle \frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$. For (c), put $x = \tan\theta$ and use the iden­tity $\displaystyle \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta$.

Step-by-step solu­tion#

Part (a)

Step 1. Inside $u = 2x$, $\displaystyle \frac{du}{dx} = 2$. Out­side $\sin^{-1}u$, deriv­a­tive $\displaystyle \frac{1}{\sqrt{1 - u^2}}$.

$$\displaystyle \frac{d}{dx}\sin^{-1}(2x) = \frac{1}{\sqrt{1 - (2x)^2}}\cdot 2 = \frac{2}{\sqrt{1 - 4x^2}}$$

Part (b)

Step 1. Inside $u = e^x$, $\displaystyle \frac{du}{dx} = e^x$. Out­side $\tan^{-1}u$, deriv­a­tive $\displaystyle \frac{1}{1 + u^2}$.

$$\displaystyle \frac{d}{dx}\tan^{-1}(e^x) = \frac{1}{1 + (e^x)^2}\cdot e^x = \frac{e^x}{1 + e^{2x}}$$

Part (c)

Step 1. Put $x = \tan\theta$. Since $0 < x < 1$, $\theta = \tan^{-1}x$ lies in $\displaystyle \left(0, \tfrac{\pi}{4}\right)$.

Step 2. Sim­plify the inside using $\displaystyle \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta$.

$$\displaystyle y = \cos^{-1}\frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos^{-1}(\cos 2\theta)$$

Step 3. Since $\displaystyle 2\theta \in \left(0, \tfrac{\pi}{2}\right)$, which lies inside $[0, \pi]$, the prin­ci­pal range of $\cos^{-1}$, we have $\cos^{-1}(\cos 2\theta) = 2\theta$.

$$y = 2\theta = 2\tan^{-1}x$$

Step 4. Dif­fer­en­ti­ate.

$$\displaystyle \frac{dy}{dx} = \frac{2}{1 + x^2}$$

Check­ing the answer#

For (c), at $x = 0.5$: a small numer­i­cal change $h = 0.001$ in $x$ changes $y$ by about $0.0016$, and $\displaystyle \frac{2}{1 + 0.25} = 1.6$. They agree.

Answer#

(a) $\displaystyle \frac{2}{\sqrt{1 - 4x^2}}$; (b) $\displaystyle \frac{e^x}{1 + e^{2x}}$; (c) $\displaystyle \frac{2}{1 + x^2}$.

Com­mon mis­take to avoid#

In (c), writ­ing $\cos^{-1}(\cos 2\theta) = 2\theta$ with­out check­ing that $2\theta$ lies in $[0, \pi]$. Here the con­di­tion $0 < x < 1$ guar­an­tees it.

Ques­tion 4: Implicit dif­fer­en­ti­a­tion#

The prob­lem#

Find $\displaystyle \frac{dy}{dx}$ for: (a) $x^3 + y^3 = 6xy$; (b) $\sin(xy) + y = x$.

Under­stand­ing the prob­lem#

In nei­ther equa­tion can $y$ be eas­ily writ­ten in terms of $x$. So you dif­fer­en­ti­ate the equa­tion as it stands, treat­ing $y$ as a func­tion of $x$.

The idea#

Dif­fer­en­ti­ate both sides with respect to $x$. Every time you dif­fer­en­ti­ate an expres­sion in $y$, mul­ti­ply by $\displaystyle y' = \frac{dy}{dx}$ (chain rule). Terms like $xy$ need the prod­uct rule. Then col­lect all $y'$ terms on one side and solve.

Step-by-step solu­tion#

Part (a)

Step 1. Dif­fer­en­ti­ate each term. $\displaystyle \frac{d}{dx}x^3 = 3x^2$; $\displaystyle \frac{d}{dx}y^3 = 3y^2y'$; $\displaystyle \frac{d}{dx}(6xy) = 6(y + xy')$ by the prod­uct rule.

$$3x^2 + 3y^2y' = 6y + 6xy'$$

Step 2. Divide by $3$ and col­lect the $y'$ terms on the left.

$$y^2y' - 2xy' = 2y - x^2$$

Step 3. Fac­tor out $y'$ and divide.

$$\displaystyle y'(y^2 - 2x) = 2y - x^2 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$$

Part (b)

Step 1. Dif­fer­en­ti­ate $\sin(xy)$: out­side $\sin$, inside $xy$, whose deriv­a­tive is $y + xy'$.

$$\displaystyle \frac{d}{dx}\sin(xy) = \cos(xy)\,(y + xy')$$

Step 2. Dif­fer­en­ti­ate the whole equa­tion.

$$\cos(xy)(y + xy') + y' = 1$$

Step 3. Expand and col­lect the $y'$ terms.

$$y\cos(xy) + x\cos(xy)\,y' + y' = 1 \quad\Longrightarrow\quad y'\bigl(1 + x\cos(xy)\bigr) = 1 - y\cos(xy)$$

Step 4. Divide.

$$\displaystyle \frac{dy}{dx} = \frac{1 - y\cos(xy)}{1 + x\cos(xy)}$$

Check­ing the answer#

For (a), the point $(3, 3)$ is on the curve ($27 + 27 = 54 = 6 \cdot 9$). The for­mula gives $\displaystyle \frac{6 - 9}{9 - 6} = -1$, and by the sym­me­try of the curve in the line $y = x$ a slope of $-1$ there is exactly what you expect. For (b), $(0, 0)$ is on the curve and the for­mula gives $\displaystyle \frac{1}{1} = 1$.

Answer#

(a) $\displaystyle \frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$; (b) $\displaystyle \frac{dy}{dx} = \frac{1 - y\cos(xy)}{1 + x\cos(xy)}$.

Com­mon mis­take to avoid#

Dif­fer­en­ti­at­ing $6xy$ as $6y'$ or $6y$. It is a prod­uct of $x$ and $y$, so the prod­uct rule gives $6(y + xy')$.

Ques­tion 5: Log­a­rith­mic dif­fer­en­ti­a­tion and vari­able pow­ers#

The prob­lem#

Find $\displaystyle \frac{dy}{dx}$ for: (a) $y = (\sin x)^x$; (b) $y = x^{\sin x}$; (c) $y = 2^x + x^2$.

Under­stand­ing the prob­lem#

In (a) and (b) both the base and the power con­tain $x$. Nei­ther the power rule ($x^n$, fixed $n$) nor the expo­nen­tial rule ($a^x$, fixed $a$) applies directly. Part (c) is a sum of two sim­ple terms.

The idea#

For vari­able pow­ers, take $\log$ of both sides first, as in Exam­ple 5, so the power comes down as a prod­uct. Then dif­fer­en­ti­ate implic­itly. For (c) use $\displaystyle \frac{d}{dx}a^x = a^x\log a$.

Step-by-step solu­tion#

Part (a)

Step 1. Take logs (on an inter­val where $\sin x > 0$).

$$\log y = x\log(\sin x)$$

Step 2. Dif­fer­en­ti­ate both sides. Left: $\displaystyle \frac{y'}{y}$. Right: prod­uct rule with $u = x$, $v = \log\sin x$, and $v' = \cot x$ (from Ques­tion 2(b)).

$$\displaystyle \frac{y'}{y} = 1\cdot\log(\sin x) + x\cot x$$

Step 3. Mul­ti­ply by $y = (\sin x)^x$.

$$\displaystyle \frac{dy}{dx} = (\sin x)^x\bigl(\log\sin x + x\cot x\bigr)$$

Part (b)

Step 1. Take logs (for $x > 0$).

$$\log y = \sin x\cdot\log x$$

Step 2. Dif­fer­en­ti­ate, using the prod­uct rule on the right.

$$\displaystyle \frac{y'}{y} = \cos x\log x + \sin x\cdot\frac{1}{x}$$

Step 3. Mul­ti­ply by $y = x^{\sin x}$.

$$\displaystyle \frac{dy}{dx} = x^{\sin x}\left(\cos x\log x + \frac{\sin x}{x}\right)$$

Part (c)

Step 1. Dif­fer­en­ti­ate term by term: $\displaystyle \frac{d}{dx}2^x = 2^x\log 2$ and $\displaystyle \frac{d}{dx}x^2 = 2x$.

$$\displaystyle \frac{dy}{dx} = 2^x\log 2 + 2x$$

Check­ing the answer#

For (b), at $x = 1$: $\log 1 = 0$, so the for­mula gives $1^{\sin 1}\cdot\sin 1 \approx 0.841$. A numer­i­cal slope of $x^{\sin x}$ at $x = 1$ also gives about $0.841$.

Answer#

(a) $(\sin x)^x(\log\sin x + x\cot x)$; (b) $\displaystyle x^{\sin x}\left(\cos x\log x + \frac{\sin x}{x}\right)$; (c) $2^x\log 2 + 2x$.

Com­mon mis­take to avoid#

Treat­ing $x^{\sin x}$ like $x^n$ and writ­ing $\sin x\cdot x^{\sin x - 1}$. That rule needs a con­stant power.

Ques­tion 6: Para­met­ric dif­fer­en­ti­a­tion#

The prob­lem#

Find $\displaystyle \frac{dy}{dx}$ for: (a) $x = t^2 + 1$, $y = t^3 - t$; (b) $x = a(\theta - \sin\theta)$, $y = a(1 - \cos\theta)$.

Under­stand­ing the prob­lem#

Here $x$ and $y$ are both given in terms of a third vari­able (a para­me­ter), $t$ or $\theta$. You want the slope $\displaystyle \frac{dy}{dx}$, which will be expressed in terms of the para­me­ter.

The idea#

Use $\displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt}$: dif­fer­en­ti­ate each with respect to the para­me­ter and divide.

Step-by-step solu­tion#

Part (a)

Step 1. Dif­fer­en­ti­ate each with respect to $t$.

$$\displaystyle \frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 3t^2 - 1$$

Step 2. Divide.

$$\displaystyle \frac{dy}{dx} = \frac{3t^2 - 1}{2t} \qquad (t \ne 0)$$

Part (b)

Step 1. Dif­fer­en­ti­ate each with respect to $\theta$.

$$\displaystyle \frac{dx}{d\theta} = a(1 - \cos\theta), \qquad \frac{dy}{d\theta} = a\sin\theta$$

Step 2. Divide; the $a$ can­cels.

$$\displaystyle \frac{dy}{dx} = \frac{a\sin\theta}{a(1 - \cos\theta)} = \frac{\sin\theta}{1 - \cos\theta}$$

Step 3. Sim­plify with the half-angle iden­ti­ties $\displaystyle \sin\theta = 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$ and $\displaystyle 1 - \cos\theta = 2\sin^2\tfrac{\theta}{2}$.

$$\displaystyle \frac{dy}{dx} = \frac{2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}}{2\sin^2\tfrac{\theta}{2}} = \cot\frac{\theta}{2}$$

Check­ing the answer#

In (b), at $\theta = \pi$ (the top of the cycloid arch) the slope is $\displaystyle \cot\tfrac{\pi}{2} = 0$, a hor­i­zon­tal tan­gent, exactly as at the high­est point of an arch.

Answer#

(a) $\displaystyle \frac{3t^2 - 1}{2t}$; (b) $\displaystyle \frac{\sin\theta}{1 - \cos\theta} = \cot\frac{\theta}{2}$.

Ques­tion 7: Sec­ond deriv­a­tives#

The prob­lem#

Find $\displaystyle \frac{d^2y}{dx^2}$ for: (a) $y = x^3\log x$; (b) $y = e^{3x}\cos x$.

Under­stand­ing the prob­lem#

The sec­ond deriv­a­tive is the deriv­a­tive of the first deriv­a­tive. Both func­tions are prod­ucts, so the prod­uct rule is needed, twice.

The idea#

Find $y'$ with the prod­uct rule, sim­plify it, then dif­fer­en­ti­ate again.

Step-by-step solu­tion#

Part (a)

Step 1. First deriv­a­tive: $u = x^3$, $v = \log x$.

$$\displaystyle y' = 3x^2\log x + x^3\cdot\frac{1}{x} = 3x^2\log x + x^2$$

Step 2. Dif­fer­en­ti­ate again. The first term is a prod­uct ($3x^2$ and $\log x$); the sec­ond is $x^2$.

$$\displaystyle y'' = 6x\log x + 3x^2\cdot\frac{1}{x} + 2x = 6x\log x + 3x + 2x$$

Step 3. Com­bine.

$$y'' = 6x\log x + 5x$$

Part (b)

Step 1. First deriv­a­tive: $u = e^{3x}$ ($u' = 3e^{3x}$), $v = \cos x$ ($v' = -\sin x$).

$$y' = 3e^{3x}\cos x - e^{3x}\sin x = e^{3x}(3\cos x - \sin x)$$

Step 2. Dif­fer­en­ti­ate again, with $u = e^{3x}$ and $w = 3\cos x - \sin x$, $w' = -3\sin x - \cos x$.

$$y'' = 3e^{3x}(3\cos x - \sin x) + e^{3x}(-3\sin x - \cos x)$$

Step 3. Take out $e^{3x}$ and col­lect: $\cos x$ terms $9 - 1 = 8$; $\sin x$ terms $-3 - 3 = -6$.

$$y'' = e^{3x}(8\cos x - 6\sin x)$$

Check­ing the answer#

For (b) at $x = 0$: $y'' = 8$. Directly, $y = e^{3x}\cos x \approx (1 + 3x + 4.5x^2)(1 - 0.5x^2) \approx 1 + 3x + 4x^2$ near $0$, whose sec­ond deriv­a­tive is $8$. Cor­rect.

Answer#

(a) $6x\log x + 5x$; (b) $e^{3x}(8\cos x - 6\sin x)$.

Ques­tion 8: Show­ing $y'' = 4y$#

The prob­lem#

If $y = e^{2x} + e^{-2x}$, show that $\displaystyle \frac{d^2y}{dx^2} = 4y$.

Under­stand­ing the prob­lem#

This is a "show that" ques­tion: com­pute $y''$ and show it equals $4$ times the orig­i­nal $y$.

The idea#

Dif­fer­en­ti­ate twice using $\displaystyle \frac{d}{dx}e^{kx} = ke^{kx}$, then com­pare with $4y$.

Step-by-step solu­tion#

Step 1. First deriv­a­tive.

$$y' = 2e^{2x} - 2e^{-2x}$$

Step 2. Sec­ond deriv­a­tive. Note $\displaystyle \frac{d}{dx}(-2e^{-2x}) = -2\cdot(-2)e^{-2x} = 4e^{-2x}$.

$$y'' = 4e^{2x} + 4e^{-2x}$$

Step 3. Fac­tor out $4$ and recog­nise $y$.

$$y'' = 4(e^{2x} + e^{-2x}) = 4y$$

Check­ing the answer#

At $x = 0$: $y = 2$ and $y'' = 8 = 4 \times 2$.

Answer#

$y'' = 4e^{2x} + 4e^{-2x} = 4y$, as required.

Ques­tion 9: A dif­fer­en­tial equa­tion sat­is­fied by $\tan^{-1}x$#

The prob­lem#

If $y = \tan^{-1}x$, show that $\displaystyle (1 + x^2)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} = 0$.

Under­stand­ing the prob­lem#

You must find $y'$ and $y''$ and sub­sti­tute them into the left side, show­ing it sim­pli­fies to $0$.

The idea#

Use $\displaystyle \frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$, then dif­fer­en­ti­ate $(1 + x^2)^{-1}$ with the chain rule.

Step-by-step solu­tion#

Step 1. First deriv­a­tive.

$$\displaystyle y' = \frac{1}{1 + x^2} = (1 + x^2)^{-1}$$

Step 2. Sec­ond deriv­a­tive by the chain rule: out­side $u^{-1}$, inside $u = 1 + x^2$ with $u' = 2x$.

$$\displaystyle y'' = -(1 + x^2)^{-2}\cdot 2x = -\frac{2x}{(1 + x^2)^2}$$

Step 3. Sub­sti­tute into the left side.

$$\displaystyle (1 + x^2)\left(-\frac{2x}{(1 + x^2)^2}\right) + 2x\cdot\frac{1}{1 + x^2} = -\frac{2x}{1 + x^2} + \frac{2x}{1 + x^2} = 0$$

Check­ing the answer#

A quicker route con­firms it: $(1 + x^2)y' = 1$; dif­fer­en­ti­at­ing both sides gives $2xy' + (1 + x^2)y'' = 0$, the same equa­tion.

Answer#

$\displaystyle y' = \frac{1}{1 + x^2}$ and $\displaystyle y'' = -\frac{2x}{(1 + x^2)^2}$, and sub­sti­tut­ing gives $(1 + x^2)y'' + 2xy' = 0$.

Ques­tion 10: Show­ing $y'' = (y')^2$#

The prob­lem#

If $e^y(x + 1) = 1$, show that $\displaystyle \frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$.

Under­stand­ing the prob­lem#

The rela­tion between $x$ and $y$ is given implic­itly, but it can be solved for $y$ eas­ily. You must find $y'$ and $y''$ and show that $y''$ equals the square of $y'$.

The idea#

Rearrange to $\displaystyle e^y = \frac{1}{x + 1}$ and take logs to get $y$ explic­itly, then dif­fer­en­ti­ate twice. (Here $x + 1 > 0$, since $e^y > 0$.)

Step-by-step solu­tion#

Step 1. Solve for $y$.

$$\displaystyle e^y = \frac{1}{x + 1} \quad\Longrightarrow\quad y = \log\frac{1}{x + 1} = -\log(x + 1)$$

Step 2. First deriv­a­tive.

$$\displaystyle y' = -\frac{1}{x + 1}$$

Step 3. Sec­ond deriv­a­tive: $\displaystyle -\frac{1}{x + 1} = -(x + 1)^{-1}$, whose deriv­a­tive is $(x + 1)^{-2}$.

$$\displaystyle y'' = \frac{1}{(x + 1)^2}$$

Step 4. Square the first deriv­a­tive.

$$\displaystyle (y')^2 = \left(-\frac{1}{x + 1}\right)^2 = \frac{1}{(x + 1)^2}$$

Step 5. The results of Steps 3 and 4 are equal, so $y'' = (y')^2$.

Check­ing the answer#

At $x = 0$: $y' = -1$ and $y'' = 1 = (-1)^2$.

Answer#

$y = -\log(x + 1)$ gives $\displaystyle y' = -\frac{1}{x + 1}$ and $\displaystyle y'' = \frac{1}{(x + 1)^2} = (y')^2$, as required.