If $y=x+e^x$, then $\frac{d^2 x}{d y^2}$ is equal to

$\frac{1}{\left(1+e^x\right)^2}$

$-\frac{e^x}{\left(1+e^x\right)^2}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

y=x+e^x

, then

\frac{d^2 x}{d y^2}

is equal to

View answer and explanation

Correct option: C

-\frac{e^x}{\left(1+e^x\right)^3}

Option C (-

\frac{e^x}{(1+e^x)^3}

) is correct because it is the result of finding the second derivative of the inverse function x(y).

Practice All Chapter MCQs →

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

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x

y

x>0

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[x]

x

x=k, k

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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}}