JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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Analyzing +4 | -1

Let

f(x)

be the function given by

f(x)=\,arc\,cos \left(\frac{1-x^2}{1+x^2}\right)

. Then,

View answer and explanation

Correct option: B

f^{\prime}(x)= \begin{cases}\frac{2}{1+x^2}, & x>0 \\ \frac{-2}{1+x^2}, & x<0\end{cases}

Option B is correct because the derivative of arccos((1-x²)/(1+x²)) is piecewise, being 2/(1+x²) for x>0 and -2/(1+x²) for x<0 due to the properties of arctangent.

More Questions from DIFFERENTIATION

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

, then

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

is equal to Let

x

and

y

be two real numbers such that

x>0

and

x y=1

. The minimum value of

x+y

is Let

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

where

[x]

denotes the integer part of

x

. Then at

x=k, k

being an integer, the left derivative of

f(x)

is If

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],

then

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

is equal to

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

is equal to If

z = y + f\left( v \right),

where

v = \left( {\frac{x}{y}} \right),

then

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}