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

, then

View answer and explanation

Correct option: A

f^{\prime}(x)=\frac{3}{1+x^2}

for all

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

Option A (

f^{\prime}(x)=\frac{{3}}{{1+x^2}}

for all

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

is correct because the derivative of the given function is

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

everywhere except at the points where the function is not defined.

More Questions from DIFFERENTIATION

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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}}