The derivative of ^{-1}(x²) w.r.t x is - CBSE Class 12 Math 2024 ...

CBSE Class 12 Math 2024 All Sets Solved Paper

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Question : 13 of 20

Marks: +1, -0

The derivative of

\tan^{-1}(x^2)

w.r.t x is

$\frac{x}{1 + x^4}$
$\frac{2x}{1 + x^4}$
$-\frac{2x}{1 + x^4}$
$\frac{1}{1 + x^4}$

Solution:

y = \tan^{-1}(x^2)

\tan y = x^2

Differentiate both sides with respect to x to get

\frac{d}{dy}(\tan y) \cdot \frac{dy}{dx} = \frac{d}{dx}(x^2)

\sec^2 y \cdot \frac{dy}{dx} = 2x

\frac{dy}{dx} = \frac{2x}{\sec^2 y}

\frac{dy}{dx} = \frac{2x}{1 + \tan^2 y} \qquad [\because \sec^2 y = 1 + \tan^2 y]

Now, replace

\tan y

with

x^2

to get

\frac{dy}{dx} = \frac{2x}{1 + x^4}

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