Question

If $y=\frac{4\sin x}{2x+\cos x},$ then $\frac{dy}{dx}=$

• A. $\frac{4(x\cos x-\sin x)+4}{(2x+\cos x{)}^2}$
• B. $\frac{8(x\cos x-\sin x)}{(2x+\cos x{)}^2}$
• C. $\frac{8(x\cos x-\sin x)+4}{(2x+\cos x{)}^2}$
• D. $\frac{4(x\cos x-\sin x)}{(2x+\cos x{)}^2}$

Answer 1

The correct option is C $\frac{8(x\cos x-\sin x)+4}{(2x+\cos x{)}^2}$

We have,
$y=\frac{4\sin x}{2x+\cos x}$
Let,
$u=4\sin x$ and $v=2x+\cos x$
$\frac{du}{dx}=4\cos x$ and $\frac{dv}{dx}=2-\sin x$
Using quotient rule:
$\frac{dy}{dx}=\frac{d\left(\frac{u}{v}\right)}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
$=\frac{(2x+\cos x)\left(4\cos x\right)-\left(4\sin x\right)(2-\sin x)}{(2x+\cos x{)}^2}$
$=\frac{8(x\cos x-\sin x)+4({\cos }^{2}x+{\sin }^{2}x)}{(2x+\cos x{)}^2}$
$=\frac{8(x\cos x-\sin x)+4}{(2x+\cos x{)}^2}$