JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

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

and

f^{\prime}(x)=\sin x^{2}

, then

\frac{d y}{d x}

View answer and explanation

Correct option: C

\frac{2\left(1+x-x^{2}\right)}{\left(x^{2}+1\right)^{2}} \sin \left(\frac{2 x-1}{x^{2}+1}\right)^{2}

The correct answer is C (

\frac{2\left(1+x-x^{2}\right)}{\left(x^{2}+1\right)^{2}} \sin \left(\frac{2 x-1}{x^{2}+1}\right)^{2}

) because it is the derivative of y with respect to x using the chain rule.

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}}