If $x \log _e\left(\log _e x\right)-x^2+y^2=4(y > 0)$, then $d y / d x$ at ${x}={e}$ is equal to:

$\frac{(1+2 {e})}{2 \sqrt{4+{e}^2}}$

$\frac{(1+2 {e})}{\sqrt{4+{e}^2}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

x \log _e\left(\log _e x\right)-x^2+y^2=4(y > 0)

, then

d y / d x

{x}={e}

is equal to:

View answer and explanation

Correct option: C

\frac{(2 {e}-1)}{2 \sqrt{4+{e}^2}}

Option C ((2e-1)/(2√(4+e²))) is correct because differentiating the given equation implicitly and substituting x=e allows us to solve for dy/dx, yielding the correct expression.

More Questions from DIFFERENTIATION

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

, then

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is equal to Let

x

and

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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),

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

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