Answer
$f'(x) = \dfrac{x^2\cos x - 2x\sin x}{(x^2+\sin x)^2}$
Work Step by Step
In order to derivate this function, you have to apply the quotient rule
$\dfrac{d}{dx}\left(\dfrac{a}{b}\right)= \dfrac{a'b-ab'}{b^2}$
So, let's identify a and b and derive them
$a = \sin x$
$a' = \cos x$
$b = x^2+\sin x$
$b' =2x+\cos x$
Then, substitute in the formula:
$f'(x) = \dfrac{(\cos x)(x^2+\sin x)-(2x+ \cos x)(\sin x)}{(x^2 +\sin x)^2}$
Simplify and get the answer:
$f'(x) = \dfrac{x^2\cos x - 2x\sin x}{(x^2+\sin x)^2}$
