<s>[INST] Simplifying the derivative of $\sin(\cos^2 x)\cos(\sin^2 x)$ Differentiating the term
$$\sin(\cos^2 x)\cos(\sin^2 x)$$
leads me through the chain and product rule to
$$-\sin(2x)\cos(\cos^2 x)\cos(\sin^2 x)+(-\sin(\sin^2 x)\sin(2x)\sin(\cos^2 x))$$
where the derivative of $\sin^2 x$ equals to
$$\frac{d}{dx} \sin^2 x = 2\sin x \frac{d}{dx} \sin x = 2\sin x \cos x = \sin 2x$$
and $\frac{d}{dx} cos^2 x$ to $-\sin 2x$ respectively.
Through factorization, I can then simplify the term to
$$-\sin 2x\ (\cos(\cos^2 x)\cos(\sin^2 x) + \sin(\sin^2 x)\sin(\cos^2 x))$$
The problem starts here where I fail to find a simplification for the second factor which, according to Wolfram Mathematica, should lead to $\cos(\cos 2x)$ and ultimately to
$$-\sin(2x)\cos(\cos 2x)$$
How and which trigonometric identities could I apply to get to that? Is my approach right?
 [/INST] Denote:
$$f(x) = \sin(\cos^2 x)\cos(\sin^2 x)$$
$$g(x) = \sin(\cos^2 x)$$
$$h(x) = \cos(\sin^2 x)$$
Thus:
$$f(x) = g(x)\cdot h(x)$$
$$g'(x) = \cos(\cos^2{x}) \cdot 2 \cos{x} \cdot (-\sin{x})$$
$$h'(x) = -\sin(\sin^2{x}) \cdot 2 \sin{x} \cdot \cos{x}$$
Now we get:
$$f'(x) = g(x)\cdot h'(x) + g'(x)\cdot h(x) $$
$$f'(x) = \sin(\cos^2 x) \cdot (-\sin(\sin^2{x}) \cdot 2 \sin{x} \cdot \cos{x}) +  \\ \cos(\cos^2{x}) \cdot 2 \cos{x} \cdot (-\sin{x}) \cdot \cos(\sin^2 x) $$
$$f'(x) = -2\sin{x}\cos{x} [ \sin(\cos^2 x) \cdot \sin(\sin^2{x}) + \cos(\cos^2 x) \cdot \cos(\sin^2{x})]$$
Now we use the formula $\cos(a-b) = \sin{a}\sin{b} + \cos{a}\cos{b}$ and we get:
$$f'(x) = -2\sin{x}\cos{x} [ \cos(\cos^2 x - \sin^2{x})]$$
$$f'(x) = -2\sin{x}\cos{x} \cos{2x}$$
$$f'(x) = -\sin{2x}\cos{2x}$$
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