How to go about proving that $\cos^2 x$ is everywhere differentiable? My first line of reasoning was to try directly evaluating $$\lim\limits_{h \to 0}\frac{\cos^2 (x+h) - \cos^2 (x)}{h}$$ and showing such a limit existed for any x, but when $\cos^2(x)$ evaluates to zero (e.g. when $x = \frac{\pi}{2}$), then directly evaluating the limit yields the indeterminate form $\frac{0}{0}$. The same thing happens even after transforming $\cos^2 x$ to $\frac{1}{2} \cos (2x)$, so I don't think this is the correct strategy. I considered using L'Hopitals rule, but don't think that is the route that the question intended, as it is suggested to use the identity $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$ along with standard trigonometric identities. I am not sure where to begin. What would be a step in the right direction?

HINT:
$$\cos^2B-\cos^2A=1-\sin^2B-(1-\sin^2A)$$
Using  Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $, this becomes  $$\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$$
So, $$\cos^2(x+h)-\cos^2x=\sin(2x+h)\sin(-h)=-\sin(2x+h)\sin(h)$$
