If $y=\sin x+e^x$, then $\frac{d^2 x}{d y^2}$ equals

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

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

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

y=\sin x+e^x

, then

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

equals

View answer and explanation

Correct option: C

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

Option C is correct because it involves finding the second derivative of x with respect to y using the chain rule and inverse function differentiation.

More Questions from DIFFERENTIATION

Practice All Chapter MCQs →

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

, then

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

is equal to Let

x

and

y

be two real numbers such that

x>0

and

x y=1

. The minimum value of

x+y

is Let

f(x)=[x] \sin (\pi x)

where

[x]

denotes the integer part of

x

. Then at

x=k, k

being an integer, the left derivative of

f(x)

is If

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],

then

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

is equal to

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

is equal to If

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

where

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

then

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