Chapter 5 · Continuity and Differentiability

Exercise 5.3 · Q13: $\frac{dy}{dx}$ if $y=\cos^{-1}\frac{2x}{1+x^2}$

Exercise 5.3Derivatives of inverse trigonometric functions · Section 5.3.3

The question

Find $\dfrac{dy}{dx}$ in the following:

y = \cos^{-1}\left(\frac{2x}{1 + x^2}\right), \quad -1 < x < 1

Exam answer

Put $x = \tan\theta$ ⁠. As $-1 < x < 1$ ⁠, $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$ ⁠, so $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$ and $0 < \frac{\pi}{2} - 2\theta < \pi$ ⁠.

Using $\sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta}$ ⁠,

\begin{aligned} y &= \cos^{-1}\left(\frac{2\tan\theta}{1 + \tan^2\theta}\right) \\ &= \cos^{-1}(\sin 2\theta) \\ &= \cos^{-1}\left(\cos\left(\frac{\pi}{2} - 2\theta\right)\right) \\ &= \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2\tan^{-1}x \end{aligned}

$\therefore\ \dfrac{dy}{dx} = -\dfrac{2}{1 + x^2}$ $(-1 < x < 1)$ ⁠.

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

The one idea this problem is built on

With $x = \tan\theta$ the fraction is $\sin 2\theta$ (as in Q9). The outer function is $\cos^{-1}$ ⁠, which undoes cosine, so we rewrite $\sin 2\theta = \cos\left(\frac{\pi}{2} - 2\theta\right)$ and check that $\frac{\pi}{2} - 2\theta$ lies in $[0, \pi]$ ⁠, the principal range of $\cos^{-1}$ ⁠. On $-1 < x < 1$ it lies in $(0, \pi)$ ⁠, so $y = \frac{\pi}{2} - 2\tan^{-1}x$ ⁠.

Given, to find, and the plan

Given: $y = \cos^{-1}\frac{2x}{1 + x^2}$ for $-1 < x < 1$ ⁠.
To find: $\dfrac{dy}{dx}$ on that interval.
Plan: (1) Put $x = \tan\theta$ and locate $\theta$ ⁠. (2) Write the fraction as $\cos\left(\frac{\pi}{2} - 2\theta\right)$ ⁠. (3) Check the range and simplify $y$ ⁠. (4) Differentiate. (5) Confirm by the chain rule directly.

Basics you need

1Principal value of $\cos^{-1}$

$\cos^{-1}: [-1, 1] \to [0, \pi]$ ⁠, and $\cos^{-1}(\cos\phi) = \phi$ exactly when $\phi \in [0, \pi]$ (Chapter 2, Sections 2.2 and 2.3). Basics: composite functions

2Two identities

$\frac{2\tan\theta}{1 + \tan^2\theta} = \sin 2\theta$ ⁠, and $\sin\alpha = \cos\left(\frac{\pi}{2} - \alpha\right)$ (Class XI). Basics: trigonometric identities

3 $\tan^{-1}$ is increasing

$\tan^{-1}(\pm1) = \pm\frac{\pi}{4}$ ⁠.

4Derivatives of $\tan^{-1}$ and $\cos^{-1}$

$\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$ ⁠; $\frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt{1 - x^2}}$ for $-1 < x < 1$ (table after Example 24). Lesson: derivatives of inverse trig functions

Solution

Substitute $x = \tan\theta$ and locate $\theta$

Let $\theta = \tan^{-1}x$ ⁠, so $x = \tan\theta$ ⁠. From $-1 < x < 1$ and the increasing function $\tan^{-1}$ ⁠,
$-\frac{\pi}{4} < \theta < \frac{\pi}{4}, \quad\text{so}\quad -\frac{\pi}{2} < 2\theta < \frac{\pi}{2}.$
The substitution works for every real $x$ ⁠; the interval for $\theta$ is what the range check in Step 3 needs.

$\tan\frac{\pi}{4} = 1$ (Chapter 2, Fig 2.5 (ii)). Basics: trigonometric identities
Write the fraction as a cosine
$\begin{aligned} \frac{2x}{1 + x^2} &= \frac{2\tan\theta}{1 + \tan^2\theta} = \sin 2\theta \\ &= \cos\left(\frac{\pi}{2} - 2\theta\right). \end{aligned}$
The first equality multiplies top and bottom by $\cos^2\theta \ne 0$ ⁠: $2\tan\theta\cos^2\theta = 2\sin\theta\cos\theta$ ⁠. The second turns the sine into a cosine, because $\cos^{-1}$ can only undo a cosine.

$\sin 2\theta = 2\sin\theta\cos\theta$ ⁠; $\cos\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha$ (Class XI). Basics: trigonometric identities
Simplify $y$ ⁠, checking the range

Multiply $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$ by $-1$ (which reverses the inequalities) and add $\frac{\pi}{2}$ ⁠:
$0 < \frac{\pi}{2} - 2\theta < \pi,$
which lies inside $[0, \pi]$ ⁠. Therefore
$\begin{aligned} y &= \cos^{-1}\left(\cos\left(\frac{\pi}{2} - 2\theta\right)\right) \\ &= \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2\tan^{-1}x. \end{aligned}$
The angle $\frac{\pi}{2} - 2\theta$ is in the principal range of $\cos^{-1}$ ⁠, so $\cos^{-1}$ returns it. The interval $-1 < x < 1$ uses the whole of that range except its endpoints.

$\cos^{-1}(\cos\phi) = \phi$ for $\phi \in [0, \pi]$ (Chapter 2, Section 2.3). Basics: composite functions
Differentiate
$\frac{dy}{dx} = 0 - 2\cdot\frac{1}{1 + x^2} = -\frac{2}{1 + x^2}.$
$\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$ (table after Example 24, Section 5.3.3). Lesson: derivatives of inverse trig functions
Cross-check with the chain rule directly

With $u = \frac{2x}{1 + x^2}$ ⁠, Q9 found $u' = \frac{2(1 - x^2)}{(1 + x^2)^2}$ and $\sqrt{1 - u^2} = \frac{|1 - x^2|}{1 + x^2}$ ⁠. For $-1 < x < 1$ ⁠, $|1 - x^2| = 1 - x^2 > 0$ ⁠, so
$\begin{aligned} \frac{dy}{dx} &= \frac{-1}{\sqrt{1 - u^2}}\cdot u' \\ &= -\frac{1 + x^2}{1 - x^2}\cdot\frac{2(1 - x^2)}{(1 + x^2)^2} = -\frac{2}{1 + x^2}. \end{aligned}$
The direct route agrees with Step 4. It also shows that $1 - x^2 > 0$ ⁠, which the interval guarantees, is exactly what keeps the sign negative; for $|x| > 1$ the derivative would be $+\frac{2}{1 + x^2}$ ⁠.

Chain rule, Theorem 4; $\frac{d}{du}\cos^{-1}u = \frac{-1}{\sqrt{1 - u^2}}$ ⁠, table after Example 24. Basics: derivative rules

For

-1 < x < 1

⁠,

\theta = \tan^{-1}x

is on the gold arc

\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)

⁠, and

\frac{\pi}{2} - 2\theta

is on the blue arc

(0, \pi)

⁠, inside the principal range

[0, \pi]

\cos^{-1}

(violet; filled dots: endpoints included).

Result

\frac{dy}{dx} = -\frac{2}{1 + x^2}, \quad -1 < x < 1.

(Textbook answer: $\dfrac{-2}{1 + x^2}$ ⁠, the same.)

Check it with numbers

At $x = 0.5$ ⁠: the fraction is $0.8$ ⁠, $y = \cos^{-1}0.8 \approx 0.6435011$ ⁠. With $h = 0.0001$ ⁠, $y(0.5001) \approx 0.6433411$ and $y(0.4999) \approx 0.6436611$ ⁠, so the difference quotient is $-1.6000000$ ⁠; the formula gives $-\frac{2}{1.25} = -1.6$ ⁠. ✓

At $x = -0.5$ ⁠: the fraction is $-0.8$ ⁠, $y = \cos^{-1}(-0.8) \approx 2.4980915$ ⁠, which equals $\frac{\pi}{2} - 2\tan^{-1}(-0.5)$ ⁠. The difference quotient is again $-1.6000000$ ⁠. ✓

Common mistakes

Writing $\cos^{-1}(\sin 2\theta) = 2\theta$ ⁠. $\cos^{-1}$ undoes cosine; rewrite $\sin 2\theta$ as $\cos\left(\frac{\pi}{2} - 2\theta\right)$ first.
Getting the inequality for $\frac{\pi}{2} - 2\theta$ backwards. Multiplying by $-1$ reverses the inequality signs.
Checking only $x > 0$ ⁠. For negative $x$ the angle $\frac{\pi}{2} - 2\theta$ is in $\left(\frac{\pi}{2}, \pi\right)$ ⁠, still in range; the check at $x = -0.5$ confirms it.

What to take away

$\cos^{-1}\frac{2x}{1 + x^2} = \frac{\pi}{2} - 2\tan^{-1}x$ for $-1 \le x \le 1$ ⁠, with derivative $-\frac{2}{1 + x^2}$ inside the interval.
Match the inner function to the outer inverse: sine inside $\sin^{-1}$ ⁠, cosine inside $\cos^{-1}$ ⁠, using $\sin\alpha = \cos\left(\frac{\pi}{2} - \alpha\right)$ to switch.

Every fact used, and where it comes from

Fact	Used in	Source
$\cos^{-1}: [-1, 1] \to [0, \pi]$ ⁠; $\tan^{-1}$ increasing	Steps 1, 3	Chapter 2, Section 2.2 and Fig 2.5 (ii)
$\cos^{-1}(\cos\phi) = \phi$ for $\phi \in [0, \pi]$	Step 3	Chapter 2, Section 2.3
$\sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta}$ ⁠; $\sin\alpha = \cos\left(\frac{\pi}{2} - \alpha\right)$	Step 2	Class XI, Trigonometric Functions
$\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$ ⁠; $\frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt{1 - x^2}}$	Steps 4, 5	Table after Example 24, Section 5.3.3 (Lesson: derivatives of inverse trig functions)
Chain rule	Step 5	Theorem 4, Section 5.3.1
Numerical values	Check	Recomputed in the site's verification script

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Exercise 5.3 · Q13: dydx\frac{dy}{dx}dxdy​ if y=cos⁡−12x1+x2y=\cos^{-1}\frac{2x}{1+x^2}y=cos−11+x22x​

Basics you need

Solution

Substitute x=tan⁡θx = \tan\thetax=tanθ and locate θ\thetaθ

Write the fraction as a cosine

Simplify yyy⁠, checking the range

Differentiate

Cross-check with the chain rule directly

Every fact used, and where it comes from

Exercise 5.3 · Q13: $\frac{dy}{dx}$ if $y=\cos^{-1}\frac{2x}{1+x^2}$

Substitute $x = \tan\theta$ and locate $\theta$

Simplify $y$ ⁠, checking the range