Question

If $y=\frac{\sin x}{x+\cos x}$, then find $\frac{dy}{dx}$.

Answer 1

Here, $u\left(x\right)=\sin x,v\left(x\right)=x+\cos x$.
$\frac{d}{dx}=\frac{(x+\cos x)\frac{d\left(\sin x\right)}{dx}-\sin x\frac{d(x+\cos x)}{dx}}{(x+\cos x{)}^2}$
$=\frac{(x+\cos x)\cos x-\cos x(1-\sin x)}{(x+\cos x{)}^2}$
$=\frac{x\cos x+{\cos }^{2}x-\sin x+{\sin }^{2}x}{(x+\cos x{)}^2}$
$=\frac{x\cos x-\sin x+{\sin }^{2}x+{\cos }^{2}x}{(x+\cos x{)}^2}$
$=\frac{x\cos x-\sin x+1}{(x+\cos x{)}^2}$.