The derivative of $f\left( {\tan x} \right){\rm{w}}{\rm{.r}}{\rm{.t}}{\rm{.g}}\left( {\sec x} \right)$ at $x = \frac{\pi }{4},$ where $f'\left( 1 \right) = 2$ and ${\rm{g'}}\left( {\sqrt 2 } \right) = 4$, is

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

The derivative of

f\left( {\tan x} \right)\mathrm{w}\mathrm{.r}\mathrm{.t}\mathrm{.g}\left( {\sec x} \right)

x = \frac{\pi }{4},

where

f'\left( 1 \right) = 2

and

\mathrm{g'}\left( {\sqrt 2 } \right) = 4

, is

View answer and explanation

Correct option: A

\frac{1}{{\sqrt 2 }}

Option A (

\frac{1}{{\sqrt 2 }}

) is correct because it is the result of differentiating f(tan x) with respect to g(sec x) and evaluating at x = pi/4.

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}}