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JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

Let

f(x)=2 /(x+1)

and

g(x)=3 x

. It is given that

(f \circ g)\left(x_0\right)=(g \circ f)\left(x_0\right)

. Then

(g \circ f)^{\prime}\left(x_0\right)

equals

View answer and explanation

Correct option: D

\frac{-32}{3}

Option D (

\frac{-32}{3}

) is correct because, after finding

x_0 = -\frac{1}{4}

, we use the chain rule to compute

(g \circ f)'(x_0) = g'(f(x_0)) * f'(x_0)

, which results in

\frac{-32}{3}

.

More Questions from DIFFERENTIATION

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

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

is equal to Let

x

y

be two real numbers such that

x>0

x y=1

. The minimum value of

x+y

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

[x]

denotes the integer part of

x

x=k, k

being an integer, the left derivative of

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}}