Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q17: $y=(\tan^{-1}x)^2$ satisfies $(x^2+1)^2y_2+2x(x^2+1)y_1=2$

Exercise 5.7Second order derivative · Section 5.7

The question

17. If $y = (\tan^{-1}x)^2$ ⁠, show that $(x^2 + 1)^2y_2 + 2x(x^2 + 1)y_1 = 2$ ⁠.

Exam answer

Here $y = (\tan^{-1}x)^2$ ⁠. By the chain rule,

\begin{aligned} y_1 &= 2\tan^{-1}x\cdot\frac{1}{1 + x^2} \\ \Rightarrow\ (1 + x^2)y_1 &= 2\tan^{-1}x \end{aligned}

Differentiating again w.r.t. $x$ (product rule on the left),

(1 + x^2)y_2 + 2xy_1 = \frac{2}{1 + x^2}

Multiplying by $1 + x^2$ ⁠,

$\therefore\ (x^2 + 1)^2y_2 + 2x(x^2 + 1)y_1 = 2$ ⁠. Hence proved.

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

The one idea this problem is built on

As in Example 38, clear the fraction before differentiating again. The first derivative is $y_1 = \frac{2\tan^{-1}x}{1+x^2}$ ⁠, so

(1 + x^2)\,y_1 = 2\tan^{-1}x.

Differentiating this with the product rule gives $(1+x^2)y_2 + 2x\,y_1 = \frac{2}{1+x^2}$ ⁠, and multiplying by $1 + x^2$ gives the relation.

Given, to show, and the plan

Given: $y = (\tan^{-1}x)^2$ ⁠, defined and differentiable for every real $x$ (because $\tan^{-1}$ is, with derivative $\frac{1}{1+x^2}$ on $\R$ ⁠). $y_1 = \frac{dy}{dx}$ ⁠, $y_2 = \frac{d^2y}{dx^2}$ ⁠.
To show: $(x^2 + 1)^2y_2 + 2x(x^2 + 1)y_1 = 2$ for every real $x$ ⁠.
Plan: (1) $y_1$ by the chain rule. (2) Clear the fraction. (3) Differentiate again with the product rule. (4) Multiply by $1 + x^2$ ⁠. (5) Independent check: compute $y_2$ explicitly and evaluate the left side term by term.

Basics you need

1Derivative of $\tan^{-1}x$

$\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ for every real $x$ (table in Section 5.3.3). Lesson: derivatives of inverse trig functions

2Chain rule for a square

$\frac{d}{dx}(u^2) = 2u\,\frac{du}{dx}$ (outer function $u^2$ ⁠, Theorem 4, Section 5.3.1). Lesson: chain rule

3Product and quotient rules

$(uv)' = u'v + uv'$ ⁠; $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ ⁠, $v \neq 0$ (Section 5.3, rules (2) and (3)). Basics: derivative rules

4Second order derivative (Section 5.7)

$y_2 = \frac{d}{dx}(y_1)$ ⁠. Lesson: second order derivative

Solution

First derivative $y_1$

Outer function $u^2$ ⁠, inner function $u = \tan^{-1}x$ ⁠. By Basics 2 and 1:
$y_1 = 2\tan^{-1}x\cdot\frac{1}{1+x^2} = \frac{2\tan^{-1}x}{1+x^2}.$
$y$ is a square of a function, so the chain rule gives "2 times the inside, times the derivative of the inside". The denominator $1 + x^2$ is at least 1, never zero, so this holds for every real $x$ ⁠.

Table of derivatives of inverse trigonometric functions, Section 5.3.3. Lesson: derivatives of inverse trig functions
Clear the fraction

Multiply both sides by $1 + x^2$ ⁠:
$(1 + x^2)\,y_1 = 2\tan^{-1}x.$
Differentiating the product on the left is simpler than differentiating a quotient, and the right side has an easy derivative. This is the trick of Example 38.
Differentiate both sides with respect to $x$

Left side, product rule with $u = 1 + x^2$ (so $u' = 2x$ ⁠) and $v = y_1$ (so $v' = y_2$ ⁠):
$\frac{d}{dx}\big[(1+x^2)\,y_1\big] = 2x\,y_1 + (1 + x^2)\,y_2.$
Right side: $\frac{d}{dx}(2\tan^{-1}x) = \frac{2}{1+x^2}$ ⁠. So
$(1 + x^2)\,y_2 + 2x\,y_1 = \frac{2}{1+x^2}.$
Both sides of Step 2 are equal for every real $x$ ⁠, so their derivatives are equal too. The derivative of $y_1$ is $y_2$ by definition.

Product rule (Basic 3); Basic 1. Basics: derivative rules
Multiply by $1 + x^2$
$(1 + x^2)^2\,y_2 + 2x(1 + x^2)\,y_1 = 2.$
Since $1 + x^2 = x^2 + 1$ ⁠, this is exactly $(x^2 + 1)^2y_2 + 2x(x^2 + 1)y_1 = 2$ ⁠, for every real $x$ ⁠.

Multiplying every term by $1 + x^2$ (never zero) clears the remaining fraction and does not change which equations are true. The result has the exact form asked for.
Independent check: $y_2$ explicitly, and the left side in full

Quotient rule on $y_1 = \frac{2\tan^{-1}x}{1+x^2}$ ⁠, with top derivative $\frac{2}{1+x^2}$ and bottom derivative $2x$ ⁠:
$\begin{aligned} y_2 &= \frac{\frac{2}{1+x^2}(1+x^2) - 2\tan^{-1}x\cdot 2x}{(1+x^2)^2} \\ &= \frac{2 - 4x\tan^{-1}x}{(1+x^2)^2}. \end{aligned}$
Now the two terms of the left side:
$\begin{aligned} (1+x^2)^2\,y_2 &= 2 - 4x\tan^{-1}x, \\ 2x(1+x^2)\,y_1 &= 2x\cdot 2\tan^{-1}x \\ &= 4x\tan^{-1}x. \end{aligned}$
Adding, the $4x\tan^{-1}x$ terms cancel and the left side is $2$ ⁠, the right side.

This route computes the left side directly, without differentiating an equation. In the first line of $y_2$ ⁠, the factor $\frac{2}{1+x^2}\cdot(1+x^2)$ simplifies to $2$ ⁠.

Quotient rule, Section 5.3, rule (3).

Result

For every real $x$ ⁠: $y_1 = \dfrac{2\tan^{-1}x}{1+x^2}$ ⁠, $y_2 = \dfrac{2 - 4x\tan^{-1}x}{(1+x^2)^2}$ ⁠, and

(x^2 + 1)^2y_2 + 2x(x^2 + 1)y_1 = 2.

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

Check it with numbers

Take $x = 1$ ⁠: $\tan^{-1}1 = \frac{\pi}{4} \approx 0.7853982$ ⁠. Then

\begin{aligned} y_1 &= \frac{2\cdot\frac{\pi}{4}}{2} = \frac{\pi}{4} \approx 0.7853982, \\ y_2 &= \frac{2 - \pi}{4} \approx -0.2853982, \end{aligned}

and the left side is exactly

4\cdot\frac{2 - \pi}{4} + 2\cdot 2\cdot\frac{\pi}{4} = 2 - \pi + \pi = 2.

From values of $y = (\tan^{-1}x)^2$ alone, with $h = 0.001$ ⁠, the second difference $\frac{y(1.001) - 2y(1) + y(0.999)}{h^2}$ is about $-0.285398$ ⁠, matching $y_2$ ⁠.

Common mistakes

Forgetting the chain rule for the square. $\frac{d}{dx}(\tan^{-1}x)^2 = 2\tan^{-1}x\cdot\frac{1}{1+x^2}$ ⁠, not $2\tan^{-1}x$ ⁠.
Differentiating only one factor of $(1+x^2)y_1$ ⁠. The product rule gives two terms, $2xy_1$ and $(1+x^2)y_2$ ⁠.
Stopping at $(1+x^2)y_2 + 2xy_1 = \frac{2}{1+x^2}$ ⁠. One more multiplication by $1 + x^2$ gives the requested form.

What to take away

Clear denominators before differentiating a second time; the product rule then produces $y_2$ and $y_1$ together, in the shape of the target relation.
The same pattern appears in Example 38 ( $y = \sin^{-1}x$ ⁠).

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ ⁠, $x \in \R$	Steps 1, 3, 5	Table after Example 24, Section 5.3.3 (Lesson: derivatives of inverse trig functions)
Chain rule: $\frac{d}{dx}u^2 = 2u\frac{du}{dx}$	Step 1	Theorem 4, Section 5.3.1 (Lesson: chain rule)
Product and quotient rules	Steps 3, 5	Section 5.3, rules (2) and (3) (Basics: derivative rules)
$y_1 = \frac{dy}{dx}$ ⁠, $y_2 = \frac{d^2y}{dx^2}$	Steps 1–5	Section 5.7 (Lesson: second order derivative)
Clearing the fraction before differentiating again	Step 2	Example 38, Section 5.7

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Exercise 5.7 · Q17: y=(tan⁡−1x)2y=(\tan^{-1}x)^2y=(tan−1x)2 satisfies (x2+1)2y2+2x(x2+1)y1=2(x^2+1)^2y_2+2x(x^2+1)y_1=2(x2+1)2y2​+2x(x2+1)y1​=2

Basics you need

Solution

First derivative y1y_1y1​

Clear the fraction

Differentiate both sides with respect to xxx

Multiply by 1+x21 + x^21+x2

Independent check: y2y_2y2​ explicitly, and the left side in full

Every fact used, and where it comes from

Exercise 5.7 · Q17: $y=(\tan^{-1}x)^2$ satisfies $(x^2+1)^2y_2+2x(x^2+1)y_1=2$

First derivative $y_1$

Differentiate both sides with respect to $x$

Multiply by $1 + x^2$

Independent check: $y_2$ explicitly, and the left side in full