JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

f(x)=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)

, then

\frac{d}{d x}(f(x))

is equal to

View answer and explanation

Correct option: C

\frac{3}{1+x^2}

for all

x \in R-\left\{-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\}

Option C is correct because it correctly simplifies the inverse tangent function and finds its derivative, considering the domain restrictions.

More Questions from DIFFERENTIATION

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}}