If $y = {e^{{{\sin }^{ - 1}}x}}$ and $u = \log x$, then $\frac{{dy}}{{du}}$ is

$\frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}$

$\frac{{{e^{{{\sin }^{ - 1}}x}}}}{x}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

y = {e^{{{\sin }^{ - 1}}x}}

and

u = \log x

, then

\frac{{dy}}{{du}}

View answer and explanation

Correct option: C

\frac{{x{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}

Option C (

\frac{{x{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}

) is correct because it is the result of finding

\frac{dy}{du}

using the chain rule.

Practice All Chapter MCQs →

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x

y

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x y=1

x+y

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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}}