JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

y = {\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }},

then

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: D

\frac{x}{{\sqrt {1 - {x^4}} }}

Option D (

\frac{x}{{\sqrt {1 - {x^4}} }}

) is correct because it is the result of simplifying the expression using trigonometric substitution and then differentiating.

More Questions from DIFFERENTIATION

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f(x)=|x|^{|\tan x|}

f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

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x y=1

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[x]

x

x=k, k

f(x)

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}