Question

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

• A. $\frac{\cos x}{2y-1}$
• B. $\frac{\sin x}{2y-1}$
• C. $\frac{\log x}{1-2y}$
• D. $\frac{\sin x}{1+xy}$

Answer 1

Answer: (A)

$y=\sqrt{\sin x+y}$
$\Rightarrow$ $y^2=sinx+y$
Differentiating w.r.t. ${}^{\prime }{x}^{\prime },$ $2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}$
$\Rightarrow$ $\frac{dy}{dx}(2y-1)=\cos x$
$\Rightarrow$ $\frac{dy}{dx}=\frac{\cos x}{2y-1}$