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 it is the result of differentiating the given equation twice with respect to x.

Practice All Chapter MCQs →

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

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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}}