If $y = {\log _u}\left| {\cos 4x} \right| + \left| {\sin x} \right|$, where $u = \sec 2x$, then $\frac{{dy}}{{dx}}$ at $x = - \frac{\pi }{6}$ is equal to

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

y = {\log _u}\left| {\cos 4x} \right| + \left| {\sin x} \right|

, where

u = \sec 2x

, then

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

x = - \frac{\pi }{6}

is equal to

View answer and explanation

Correct option: A

\frac{{ - 6\sqrt 3 }}{{{{\log }_e}2}} - \frac{{\sqrt 3 }}{2}

Option A (

\frac{{ - 6\sqrt 3 }}{{{{\log }_e}2}} - \frac{{\sqrt 3 }}{2}

) is correct because it is the result of differentiating the given function and substituting

x = -\frac{\pi}{6}

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}}