Question

If $siny=xsin(a+y)$, then $\frac{dy}{dx}$ is

• A. $\frac{sina}{si{n}^2(a+y)}$
• B. $\frac{si{n}^2(a+y)}{sina}$
• C. $sinasi{n}^2(a+y)$
• D. $\frac{si{n}^2(a-y)}{sina}$

Answer 1

The correct option is B $\frac{si{n}^2(a+y)}{sina}$

Given,

$\sin y=x\sin (a+y)$

$x=\frac{\sin y}{\sin (a+y)}$

Formula:$\frac{d}{dx}\left(\frac{u}{v}\right)$ $=\frac{u\frac{d}{dx}\left(v\right)-\left(v\right)\frac{d}{dx}\left(u\right)}{v^2}$

$\frac{dx}{dy}=\frac{\sin (a+y)\frac{d}{dy}\sin y-\sin y\frac{d}{dx}\sin (a+y)}{{\sin }^2(a+y)}$

$=\frac{\sin (a+y)\cos y-\sin y\cos (a+y)}{{\sin }^2(a+y)}$

$=\frac{\sin \left[\right(a+y)-y]}{{\sin }^2(a+y)}$

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

$\frac{dy}{dx}=\frac{{\sin }^2(a+y)}{\sin y}$