## Calculus, 10th Edition (Anton)

$f'(x) = \dfrac{x^2\cos x - 2x\sin x}{(x^2+\sin x)^2}$
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: $f'(x) = \dfrac{x^2\cos x - 2x\sin x}{(x^2+\sin x)^2}$