If $\sin y = x\sin \left( {a + y} \right),$ then $\frac{{dy}}{{dx}}$ is

$\frac{{\sin a}}{{{{\sin }^2}\left( {a + y} \right)}}$

$\sin a{\sin ^2}\left( {a + y} \right)$

$\frac{{{\rm{si}}{{\rm{n}}^2}\left( {a - {\rm{y}}} \right)}}{{\sin a}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

\sin y = x\sin \left( {a + y} \right),

then

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

View answer and explanation

Correct option: B

\frac{{{{\sin }^2}\left( {a + y} \right)}}{{\sin a}}

Option B (

\frac{{{{\sin }^2}\left( {a + y} \right)}}{{\sin a}}

) is correct because it is the result of differentiating the given equation and solving for

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

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}}