The derivative of \tan^{-1} (x^2) w.r.t x is

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

$\;{x}/{1 + x^4}$
$\;{2x}/{1 + x^4}$
$-\;{2x}/{1 + x^4}$
$\;{1}/{1 + x^4}$

Solution:

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

\tan y= x^2

Differentiate both sides with respect to x to get

\;{d}/{dy}(\tan y). \;{dy}/{dx} = \;{d}/{dx} (x^2)

\sec^2 y. \;{dy}/{dx} = 2x

\;{dy}/{dx} = \;{2x}/{\sec^2 y}

\;{dy}/{dx} = \;{2x}/{1 + \tan^2 y} \;\; [ ∵ \sec^2y = 1 + \tan^2 y]

Now, replace

\tan y

with

x^2

to get

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

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