If $ \sin (x y)+\frac{x}{y}=x^{2}-y $, then $ \frac{d y}{d x} $ is equal to

$ \frac{\left[2 x y-y^{2} \cos (x y)-1\right]}{x y^{2} \cos (x y)+y^{2}-x} $

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

\sin (x y)+\frac{x}{y}=x^{2}-y

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

\frac{y\left[2 x y-y^{2} \cos (x y)-1\right]}{x y^{2} \cos (x y)+y^{2}-x}

Option A (

\frac{y[2 x y-y^{2} \cos (x y)-1]}{x y^{2} \cos (x y)+y^{2}-x}

) is correct because it is the result of implicit 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}}