If $y = {e^x}.{e^{{x^2}}}.{e^{{x^3}}}{._.}...{e^{{x^n}}}...,$ for $0 < x < 1$, then $\frac{{dy}}{{dx}}$ at $x = \frac{1}{2}$ is

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

y = {e^x}.{e^{{x^2}}}.{e^{{x^3}}}{._.}...{e^{{x^n}}}...,

for

0 < x < 1

, then

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

x = \frac{1}{2}

View answer and explanation

Correct option: B

4e

Option B (

4e

) is correct because after simplifying the expression and differentiating, the derivative evaluated at

x = \frac{1}{2}

4e

Practice All Chapter MCQs →

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

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

x

y

x>0

x y=1

x+y

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

[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}}