Question

If $y=\frac{\sin 4x}{x^2+16}$, then find $\frac{dy}{dx}$

Answer 1

Given that:

$y=\frac{\sin 4x}{x^2+16}$

Differentiate with respect to x,

$\frac{dy}{dx}=\frac{d\left(\frac{\sin 4x}{x^2+16}\right)}{dx}$

$\frac{dy}{dx}=\frac{\cos 4x\times 4\times (x^2+16)-\sin 4x\left(2x\right)}{(x^2+16{)}^2}$

$\frac{dy}{dx}=\frac{4\cos 4x(x^2+16-2x\sin 4x)}{(x^2+16{)}^2}$