Question

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

• A. $\frac{\sin a}{{\sin }^2(a+y)}$
• B. $\frac{{\sin }^2(a+y)}{\sin a}$
• C. $\sin a{\sin }^2(a+y)$
• D. $\frac{{\sin }^2(a-y)}{\sin a}$

Answer 1

The correct option is B

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

Explanation for the correct option:

Step 1: Rearrange the given function, then differentiate both the sides of the given function with respect to x, Using the Chain rule that is the derivative of $f\left(g\right(x\left)\right)$ is $f\text{'}\left(g\right(x\left)\right)\cdot g\text{'}\left(x\right)$.

$\begin{array}{c}\sin y=x\sin (a+y)\\ \frac{\sin y}{\sin (a+y)}=x\end{array}$

Differentiating both sides with respect to $x$,

$\begin{array}{c}\frac{\cos y\frac{dy}{dx}\sin (a+y)-\cos (a+y)\frac{dy}{dx}\sin y}{{\sin }^2(a+y)}=1\\ \frac{\cos y\sin (a+y)-\cos (a+y)\sin y}{{\sin }^2(a+y)}\left(\frac{dy}{dx}\right)=1\\ \frac{\sin (a+y-y)}{{\sin }^2(a+y)}\left(\frac{dy}{dx}\right)=1\left(\because \sin (a-b)=\sin a\cos b-\cos a\sin b\right)\\ \frac{\sin a}{{\sin }^2(a+y)}\left(\frac{dy}{dx}\right)=1\\ \frac{dy}{dx}=\frac{{\sin }^2(a+y)}{\sin a}\end{array}$

Hence, the correct option is (B).