Product rule with sine and cosine

Apply the product rule with $f(x) = \sin(x)$ and $g(x) = \cos(x)$. Since $f'(x) = \cos(x)$ and $g'(x) = -\sin(x)$, the product rule gives $h'(x) = f'(x)g(x) + f(x)g'(x) = \cos(x)\cos(x) + \sin(x)(-\sin(x))$.

This simplifies to $h'(x) = \cos^2(x) - \sin^2(x)$. Using the double angle identity, this can also be written as $h'(x) = \cos(2x)$, but either form is an acceptable simplified answer.

The negative sign belongs to sine squared. Recall from part a that applying the product rule to sine times cosine gave cosine squared minus sine squared, since the derivative of cosine is negative sine. This negative sine, multiplied by sine, produces the negative sine squared term.