Chapter 5 · Continuity and Differentiability

Exercise 5.3 · Q14: $\frac{dy}{dx}$ if $y=\sin^{-1}(2x\sqrt{1-x^2})$

Exercise 5.3Derivatives of inverse trigonometric functions · Section 5.3.3

The question

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

y = \sin^{-1}\left(2x\sqrt{1 - x^2}\right), \quad -\frac{1}{\sqrt2} < x < \frac{1}{\sqrt2}

Exam answer

Put $x = \sin\theta$ ⁠. As $-\frac{1}{\sqrt2} < x < \frac{1}{\sqrt2}$ ⁠, $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$ ⁠, so $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$ and $\cos\theta > 0$ ⁠.

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

$\therefore\ \dfrac{dy}{dx} = \dfrac{2}{\sqrt{1 - x^2}}$ $\left(-\dfrac{1}{\sqrt2} < x < \dfrac{1}{\sqrt2}\right)$ ⁠.

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

The one idea this problem is built on

The expression $2x\sqrt{1 - x^2}$ looks like $2\sin\theta\cos\theta = \sin 2\theta$ with $x = \sin\theta$ ⁠. Two checks make this rigorous: $\sqrt{1 - x^2}$ equals $\cos\theta$ (not $-\cos\theta$ ⁠) because $\cos\theta > 0$ ⁠, and $2\theta$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ⁠, so $\sin^{-1}(\sin 2\theta) = 2\theta$ ⁠. The interval $|x| < \frac{1}{\sqrt2}$ is exactly what gives $\theta \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ ⁠. This is Example 3(i) of Chapter 2, now differentiated.

Given, to find, and the plan

Given: $y = \sin^{-1}\left(2x\sqrt{1 - x^2}\right)$ for $-\frac{1}{\sqrt2} < x < \frac{1}{\sqrt2}$ ⁠.
To find: $\dfrac{dy}{dx}$ on that interval.
Plan: (1) Put $x = \sin\theta$ and locate $\theta$ ⁠. (2) Show $\sqrt{1 - x^2} = \cos\theta$ ⁠, so the inside is $\sin 2\theta$ ⁠. (3) Check $2\theta$ is in range and simplify $y$ ⁠. (4) Differentiate. (5) Confirm by the chain rule directly.

Basics you need

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

$\sin^{-1}: [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ⁠, increasing; $\sin(\sin^{-1}x) = x$ for $x \in [-1, 1]$ ⁠, and $\sin^{-1}(\sin\phi) = \phi$ for $\phi \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (Chapter 2, Sections 2.2 and 2.3). Basics: composite functions

2Square roots

$\sqrt{a^2} = |a|$ ⁠; it equals $a$ only when $a \ge 0$ ⁠.

3Identities

$\sin^2\theta + \cos^2\theta = 1$ ⁠, $2\sin\theta\cos\theta = \sin 2\theta$ ⁠; $\cos\theta > 0$ for $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ (Class XI). Basics: trigonometric identities

4Derivative of $\sin^{-1}x$

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

Solution

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

Let $\theta = \sin^{-1}x$ ⁠. This is defined because $|x| < \frac{1}{\sqrt2} < 1$ ⁠, and it gives $\sin\theta = x$ ⁠. Applying the increasing function $\sin^{-1}$ to $-\frac{1}{\sqrt2} < x < \frac{1}{\sqrt2}$ ⁠, with $\sin^{-1}\frac{1}{\sqrt2} = \frac{\pi}{4}$ ⁠:
$-\frac{\pi}{4} < \theta < \frac{\pi}{4}, \quad\text{so}\quad -\frac{\pi}{2} < 2\theta < \frac{\pi}{2}.$
The substitution $x = \sin\theta$ is legitimate only for $|x| \le 1$ ⁠; here that holds. The interval for $\theta$ is needed twice: for the sign of $\cos\theta$ (Step 2) and for the range of $2\theta$ (Step 3).

$\sin\frac{\pi}{4} = \frac{1}{\sqrt2}$ ⁠; $\sin^{-1}$ is increasing (Chapter 2, Fig 2.1). Basics: trigonometric identities
The square root is $\cos\theta$ ⁠, so the inside is $\sin 2\theta$
$\sqrt{1 - x^2} = \sqrt{1 - \sin^2\theta} = \sqrt{\cos^2\theta} = |\cos\theta| = \cos\theta,$
because $\cos\theta > 0$ for $\theta \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ ⁠. Hence
$2x\sqrt{1 - x^2} = 2\sin\theta\cos\theta = \sin 2\theta.$
This is the step most often done carelessly. $\sqrt{\cos^2\theta}$ is $|\cos\theta|$ ⁠; dropping the absolute value is correct only because $\theta$ lies where cosine is positive. (Any $\theta$ in the principal range of $\sin^{-1}$ would do for this step.)

$\sin^2\theta + \cos^2\theta = 1$ ⁠; $\sqrt{a^2} = |a|$ (Class XI). Basics: trigonometric identities
Simplify $y$ ⁠, checking the range

Since $2\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ⁠, inside the principal range of $\sin^{-1}$ ⁠,
$y = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x.$
Here the interval really matters. For $\frac{1}{\sqrt2} < x < 1$ ⁠, $2\theta$ would exceed $\frac{\pi}{2}$ ⁠, and Chapter 2, Example 3(ii) shows that then $y = 2\cos^{-1}x$ instead.

$\sin^{-1}(\sin\phi) = \phi$ for $\phi \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (Chapter 2, Section 2.3; Example 3(i)). Basics: composite functions
Differentiate
$\frac{dy}{dx} = 2\cdot\frac{1}{\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}}.$
The formula for $\frac{d}{dx}\sin^{-1}x$ needs $-1 < x < 1$ ⁠, which holds on our interval.

Example 24 and the table after it, Section 5.3.3. Lesson: derivatives of inverse trig functions
Cross-check with the chain rule directly

Let $u = 2x\sqrt{1 - x^2}$ ⁠. By the product and chain rules,
$\begin{aligned} u' &= 2\sqrt{1 - x^2} + 2x\cdot\frac{-x}{\sqrt{1 - x^2}} \\ &= \frac{2(1 - x^2) - 2x^2}{\sqrt{1 - x^2}} \\ &= \frac{2(1 - 2x^2)}{\sqrt{1 - x^2}}. \end{aligned}$
Also $1 - u^2 = 1 - 4x^2 + 4x^4$ $= (1 - 2x^2)^2$ ⁠, and $1 - 2x^2 > 0$ because $x^2 < \frac12$ ⁠, so $\sqrt{1 - u^2} = 1 - 2x^2$ ⁠. Then
$\begin{aligned} \frac{dy}{dx} &= \frac{u'}{\sqrt{1 - u^2}} \\ &= \frac{2(1 - 2x^2)}{\sqrt{1 - x^2}\,(1 - 2x^2)} \\ &= \frac{2}{\sqrt{1 - x^2}}. \end{aligned}$
The direct route agrees with Step 4. It also shows the role of the interval: if $x^2 > \frac12$ ⁠, then $\sqrt{1 - u^2} = 2x^2 - 1$ and the sign flips; at $x^2 = \frac12$ ⁠, $u = \pm1$ and there is no derivative.

Product rule (2), Section 5.3; chain rule, Theorem 4; $\frac{d}{dx}\sqrt{1 - x^2} = \frac{-x}{\sqrt{1 - x^2}}$ ⁠. Basics: derivative rules

For

|x| < \frac{1}{\sqrt2}

⁠,

\theta = \sin^{-1}x

is on the gold arc

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

⁠, where

\cos\theta > 0

⁠, and

2\theta

is on the blue arc

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

⁠, inside the principal range of

\sin^{-1}

(violet; filled dots: endpoints included).

Result

\frac{dy}{dx} = \frac{2}{\sqrt{1 - x^2}}, \quad -\frac{1}{\sqrt2} < x < \frac{1}{\sqrt2}.

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

Check it with numbers

At $x = 0.5$ (allowed, since $\frac{1}{\sqrt2} \approx 0.7071$ ⁠): $2x\sqrt{1 - x^2} = \sqrt{0.75} \approx 0.8660254$ ⁠, so $y = \sin^{-1}0.8660254$ ⁠, which is $\frac{\pi}{3} \approx 1.0471976$ ⁠, and $2\sin^{-1}0.5 = \frac{\pi}{3}$ too. With $h = 0.0001$ ⁠, $y(0.5001) \approx 1.0474285$ and $y(0.4999) \approx 1.0469666$ ⁠, and the difference quotient $\frac{y(0.5001) - y(0.4999)}{0.0002}$ ⁠, computed from the unrounded values, is $2.3094011$ ⁠; the formula gives $\frac{2}{\sqrt{0.75}} \approx 2.3094011$ ⁠. ✓

Common mistakes

Writing $\sqrt{1 - \sin^2\theta} = \cos\theta$ without a reason. It is $|\cos\theta|$ ⁠; it equals $\cos\theta$ here because $\theta \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ ⁠.
Skipping the range of $2\theta$ ⁠. Without $|x| < \frac{1}{\sqrt2}$ ⁠, $2\theta$ can leave $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\sin^{-1}(\sin 2\theta) \ne 2\theta$ ⁠.
Substituting $x = \cos\theta$ without adjusting. That route leads to $\sin^{-1}(\sin 2\theta)$ with $2\theta \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$ ⁠, which needs the reflection $\pi - 2\theta$ and more care.

What to take away

$\sin^{-1}\left(2x\sqrt{1 - x^2}\right) = 2\sin^{-1}x$ for $|x| \le \frac{1}{\sqrt2}$ (Chapter 2, Example 3(i)), so its derivative is $\frac{2}{\sqrt{1 - x^2}}$ inside that interval.
With $x = \sin\theta$ ⁠, always settle the sign of $\sqrt{1 - x^2} = |\cos\theta|$ from the interval for $\theta$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin^{-1}: [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ⁠, increasing	Steps 1, 3	Chapter 2, Section 2.2 and Fig 2.1
$\sin^{-1}(\sin\phi) = \phi$ for $\phi \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ⁠; $\sin^{-1}(2x\sqrt{1 - x^2}) = 2\sin^{-1}x$	Step 3	Chapter 2, Section 2.3 and Example 3
$\sin^2\theta + \cos^2\theta = 1$ ⁠; $2\sin\theta\cos\theta = \sin 2\theta$ ⁠; sign of cosine	Step 2	Class XI, Trigonometric Functions
$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$ ⁠, $-1 < x < 1$	Steps 4, 5	Example 24, Section 5.3.3 (Lesson: derivatives of inverse trig functions)
Product rule; chain rule	Step 5	Rule (2), Section 5.3; Theorem 4, Section 5.3.1
Numerical values	Check	Recomputed in the site's verification script

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Exercise 5.3 · Q14: dydx\frac{dy}{dx}dxdy​ if y=sin⁡−1(2x1−x2)y=\sin^{-1}(2x\sqrt{1-x^2})y=sin−1(2x1−x2​)

Basics you need

Solution

Substitute x=sin⁡θx = \sin\thetax=sinθ and locate θ\thetaθ

The square root is cos⁡θ\cos\thetacosθ⁠, so the inside is sin⁡2θ\sin 2\thetasin2θ

Simplify yyy⁠, checking the range

Differentiate

Cross-check with the chain rule directly

Every fact used, and where it comes from

Exercise 5.3 · Q14: $\frac{dy}{dx}$ if $y=\sin^{-1}(2x\sqrt{1-x^2})$

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

The square root is $\cos\theta$ ⁠, so the inside is $\sin 2\theta$

Simplify $y$ ⁠, checking the range