Question

If $y=(x^2+1{)}^5\sin 4x$, find $\frac{dy}{dx}$.

Answer 1

$\frac{dy}{dx}=5(x^2+1).2x.sin4x+4cos4x(x^2+1)$

$\frac{dy}{dx}=10x(x^2+1)sin4x+4cos4x(x^2+1)$