Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q16: $e^y(x+1)=1$ gives $y''=(y')^2$

Exercise 5.7Second order derivative · Section 5.7

The question

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

Exam answer

Here $e^y(x + 1) = 1$ ⁠, so $x + 1 = e^{-y} \gt 0$ ⁠, i.e. $x \gt -1$ ⁠.

Differentiating both sides w.r.t. $x$ ⁠,

\begin{aligned} e^y\frac{dy}{dx}(x + 1) + e^y &= 0 \\ \Rightarrow\ \frac{dy}{dx} &= -\frac{1}{x + 1} \end{aligned}

Differentiating again w.r.t. $x$ ⁠,

\begin{aligned} \frac{d^2y}{dx^2} &= -(-1)(x + 1)^{-2} \\ &= \frac{1}{(x + 1)^2} \\ &= \Big(-\frac{1}{x + 1}\Big)^2 \end{aligned}

$\therefore\ \dfrac{d^2y}{dx^2} = \Big(\dfrac{dy}{dx}\Big)^2$ ⁠. Hence proved.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Differentiating the equation once gives $\frac{dy}{dx} = -\frac{1}{x+1}$ ⁠, and the equation itself says $\frac{1}{x+1} = e^y$ ⁠. So

\frac{dy}{dx} = -e^y.

Differentiating once more gives $\frac{d^2y}{dx^2} = -e^y\cdot\frac{dy}{dx}$ ⁠, and the factor $-e^y$ is $\frac{dy}{dx}$ again. So the second derivative is the first derivative times itself.

Given, to show, and the plan

Given: $e^y(x + 1) = 1$ ⁠, which defines $y$ as a function of $x$ ⁠.
To show: $\dfrac{d^2y}{dx^2} = \left(\dfrac{dy}{dx}\right)^2$ wherever the relation defines $y$ ⁠.
Plan: (1) Find the domain: which $x$ are possible, and solve for $y$ ⁠. (2) Differentiate the equation implicitly to get $\frac{dy}{dx}$ ⁠. (3) Differentiate again. (4) Compute both sides in full as functions of $x$ and compare. (5) Check with the explicit formula.

Basics you need

1Exponential and logarithm

$e^y \gt 0$ for every real $y$ ⁠; $\log$ is defined for positive numbers only; $e^y = c$ (with $c \gt 0$ ⁠) means $y = \log c$ ⁠; $\frac{d}{dx}\log x = \frac{1}{x}$ (Section 5.4). Basics: exponentials and logarithms

2Implicit differentiation

When $y$ is a function of $x$ ⁠, $\frac{d}{dx}g(y) = g'(y)\,\frac{dy}{dx}$ ⁠. In particular $\frac{d}{dx}e^y = e^y\,\frac{dy}{dx}$ (Section 5.3.2). Lesson: implicit differentiation

3Product rule

$(uv)' = u'v + uv'$ (Section 5.3, rule (2)). Basics: derivative rules

4Second order derivative (Section 5.7)

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ ⁠. Lesson: second order derivative

Solution

Domain: $x \gt -1$ ⁠, and $y = -\log(x+1)$

Since $e^y \gt 0$ ⁠, the equation $e^y(x+1) = 1$ gives $x + 1 = \frac{1}{e^y} = e^{-y} \gt 0$ ⁠, so $x \gt -1$ ⁠. Conversely, for each $x \gt -1$ the equation $e^y = \frac{1}{x+1}$ has exactly one solution,
$y = \log\frac{1}{x+1} = -\log(x+1).$
So the relation defines $y$ as a differentiable function of $x$ on $x \gt -1$ ⁠, and we work there.

Before differentiating an implicit relation it is worth knowing where it actually defines a function. Here $x \le -1$ is impossible, because the left side would be zero or negative. Knowing the explicit form also guarantees that $y$ is differentiable, which implicit differentiation assumes.

$e^y \gt 0$ ⁠; $\log\frac{1}{c} = -\log c$ ⁠. Basics: exponentials and logarithms
First derivative by implicit differentiation

Differentiate both sides of $e^y(x+1) = 1$ with respect to $x$ ⁠. Product rule on the left, with $u = e^y$ and $v = x + 1$ ⁠:
$e^y\frac{dy}{dx}\cdot(x+1) + e^y\cdot 1 = 0.$
Divide by $e^y$ (never zero) and solve, using $x + 1 \neq 0$ ⁠:
$(x+1)\frac{dy}{dx} = -1, \qquad \frac{dy}{dx} = -\frac{1}{x+1}.$
$e^y$ depends on $x$ through $y$ ⁠, so its derivative is $e^y\frac{dy}{dx}$ (Basic 2). The right side is the constant 1, with derivative 0. Dividing by $e^y$ and by $x + 1$ is safe because neither is zero on the domain.

Basics 2 and 3. Lesson: implicit differentiation
Rewrite $\frac{dy}{dx}$ using the equation

From $e^y(x+1) = 1$ ⁠, $\frac{1}{x+1} = e^y$ ⁠. So
$\frac{dy}{dx} = -e^y.$
This form is in $y$ only, and it is very easy to differentiate again. It is also where the square in the target comes from, as the next step shows.
Second derivative

Differentiate $\frac{dy}{dx} = -e^y$ with respect to $x$ (Basic 2):
$\frac{d^2y}{dx^2} = -e^y\cdot\frac{dy}{dx}.$
By Step 3, $-e^y = \frac{dy}{dx}$ ⁠. Substituting,
$\frac{d^2y}{dx^2} = \frac{dy}{dx}\cdot\frac{dy}{dx} = \left(\frac{dy}{dx}\right)^2.$
The derivative of $-e^y$ with respect to $x$ is $-e^y$ times $\frac{dy}{dx}$ (chain rule). The factor $-e^y$ is itself $\frac{dy}{dx}$ ⁠, so the second derivative is the square of the first. This proves the relation for every $x \gt -1$ ⁠.

Chain rule (Theorem 4, Section 5.3.1) and Section 5.7. Lesson: second order derivative
Both sides in full, as functions of $x$

Left side: differentiate $\frac{dy}{dx} = -(x+1)^{-1}$ by the power rule with the chain rule (inner $x + 1$ ⁠, derivative 1):
$\frac{d^2y}{dx^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2}.$
Right side:
$\left(\frac{dy}{dx}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}.$
The two sides are equal for every $x \gt -1$ ⁠.

This computes each side completely, independently of the trick in Step 4. Squaring removes the minus sign, which matches the positive second derivative.

$\frac{d}{dx}x^n = nx^{n-1}$ with $n = -1$ (Table 5.3).
Check with the explicit formula

With $y = -\log(x+1)$ from Step 1, the chain rule gives
$\frac{dy}{dx} = -\frac{1}{x+1}\cdot 1 = -\frac{1}{x+1},$
the same as Step 2, and then $\frac{d^2y}{dx^2} = \frac{1}{(x+1)^2}$ as in Step 5.

Implicit and explicit differentiation must agree; when the explicit form is easy to find, it is a quick independent check.

$\frac{d}{dx}\log x = \frac{1}{x}$ ⁠, Theorem 5, Section 5.4. Basics: exponentials and logarithms

The curve

e^y(x+1) = 1

⁠, that is

y = -\log(x+1)

for

x \gt -1

(equal scales). It falls everywhere (

\frac{dy}{dx} \lt 0

⁠) and bends upward everywhere (

\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2 \gt 0

⁠). The line

x = -1

is never reached. The marked point

(1, -\log 2)

is used in the numerical check.

Result

The relation $e^y(x+1) = 1$ means $y = -\log(x+1)$ ⁠, $x \gt -1$ ⁠. There $\dfrac{dy}{dx} = -e^y = -\dfrac{1}{x+1}$ and

\frac{d^2y}{dx^2} = \frac{1}{(x+1)^2} = \left(\frac{dy}{dx}\right)^2.

(The answer key gives no entry for this "show that" question.)

Check it with numbers

Take $x = 1$ ⁠: $y = -\log 2 \approx -0.6931472$ ⁠, and indeed $e^y(x+1) = \frac12 \cdot 2 = 1$ ⁠. Then $\frac{dy}{dx} = -\frac12$ ⁠, $\left(\frac{dy}{dx}\right)^2 = \frac14$ and $\frac{d^2y}{dx^2} = \frac{1}{2^2} = \frac14$ ⁠. Also $-e^y = -\frac12$ ⁠, as Step 3 says.

From values of $y = -\log(x+1)$ alone, with $h = 0.001$ ⁠, the second difference $\frac{y(1.001) - 2y(1) + y(0.999)}{h^2}$ is about $0.250000$ ⁠.

Common mistakes

Differentiating $e^y$ as $e^y$ ⁠. With respect to $x$ it is $e^y\frac{dy}{dx}$ ⁠.
Forgetting the product rule. The left side $e^y(x+1)$ is a product of two functions of $x$ ⁠.
Not stating the domain. The relation only makes sense for $x \gt -1$ ⁠.
Writing $\frac{d^2y}{dx^2} = -e^y$ ⁠. Differentiating $-e^y$ with respect to $x$ needs the factor $\frac{dy}{dx}$ ⁠.

What to take away

After implicit differentiation, use the original equation to simplify $\frac{dy}{dx}$ (here to $-e^y$ ⁠); the simplified form is often the key to the second derivative.
For a "show that" relation, compute both sides fully; here both equal $\frac{1}{(x+1)^2}$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$e^y \gt 0$ ⁠; $\log$ defined for positive numbers; $\frac{d}{dx}\log x = \frac{1}{x}$ ⁠, $\frac{d}{dx}e^x = e^x$	Steps 1, 2, 6	Section 5.4, Theorem 5 (Lesson: exponential and logarithmic functions)
Implicit differentiation: $\frac{d}{dx}e^y = e^y\frac{dy}{dx}$	Steps 2, 4	Section 5.3.2, with the chain rule, Theorem 4 (Lesson: implicit differentiation)
Product rule	Step 2	Section 5.3, rule (2) (Basics: derivative rules)
$\frac{d}{dx}x^n = nx^{n-1}$	Step 5	Table 5.3, Section 5.3
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Steps 4, 5	Section 5.7 (Lesson: second order derivative)

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Exercise 5.7 · Q16: ey(x+1)=1e^y(x+1)=1ey(x+1)=1 gives y′′=(y′)2y''=(y')^2y′′=(y′)2

Basics you need

Solution

Domain: x>−1x \gt -1x>−1⁠, and y=−log⁡(x+1)y = -\log(x+1)y=−log(x+1)

First derivative by implicit differentiation

Rewrite dydx\frac{dy}{dx}dxdy​ using the equation

Second derivative

Both sides in full, as functions of xxx

Check with the explicit formula

Every fact used, and where it comes from

Exercise 5.7 · Q16: $e^y(x+1)=1$ gives $y''=(y')^2$

Domain: $x \gt -1$ ⁠, and $y = -\log(x+1)$

Rewrite $\frac{dy}{dx}$ using the equation

Both sides in full, as functions of $x$