If $ \tan ^{-1} y-y+x=0 $ then $ \frac{d^{2} y}{d x^{2}} $ is equal to

$ \frac{2\left(1+y^{2}\right)}{y^{4}} $

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

\tan ^{-1} y-y+x=0

then

\frac{d^{2} y}{d x^{2}}

is equal to

View answer and explanation

Correct option: A

\frac{-2\left(1+y^{2}\right)}{y^{5}}

Option A (

\frac{{-2(1+y^{2})}}{{y^{5}}}

) is correct because differentiating the given equation twice and solving for

\frac{{d^{2} y}}{{d x^{2}}}

yields this result.

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}}