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?
 A: Use https://mathworld.wolfram.com/WernerFormulas.html,
$$2\sin(\cos^2x)\cos(\sin^2x)=\sin(1)+\sin(\cos2x)$$
before differentiation
A: You are correct.
Using the identities $\cos(A)\cos(B)+\sin(A)\sin(B)=\cos(A-B)$ and $\cos(2A)=\cos^2(A)-\sin^2(A)$ we have $$\frac{d}{dx}(\sin(\cos^2 x)\cos(\sin^2 x))$$
$$=-\sin(2x)\big(\cos(\cos^2 x)\cos(\sin^2 x)+\sin(\cos^2 x)\sin(\sin^2 x)\big)$$
$$=-\sin(2x)\cos(\cos^2(x)-\sin^2(x))$$
$$=-\sin(2x)\cos(\cos(2x))$$
A: 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:
$$f'(x) = -2\sin{x}\cos{x} [ \cos(\cos^2 x - \sin^2{x})]$$
$$f'(x) = -2\sin{x}\cos{x} \cos{2x}$$
$$f'(x) = -\sin{2x}\cos{2x}$$
A: If we start with $$y=\sin \left(f(x)^2\right) \cos \left(g(x)^2\right)$$
$$y'=2 f(x) f'(x) \cos \left(f(x)^2\right) \cos \left(g(x)^2\right)-2 g(x) g'(x)\sin
   \left(f(x)^2\right)  \sin \left(g(x)^2\right)$$ If $f(x)=\cos(x)$ and $g(x)=\sin(x)$, then
$$2 f(x) f'(x)=-2 \sin (x) \cos (x)=-\sin(2x)$$
$$2 g(x) g'(x)=+2 \sin (x) \cos (x)=+\sin(2x)$$ So
$$y'=-\sin(2x)\Big[\cos \left(f(x)^2\right) \cos \left(g(x)^2\right)+\sin
   \left(f(x)^2\right)  \sin \left(g(x)^2\right) \Big]$$
$$\cos \left(f(x)^2\right) \cos \left(g(x)^2\right)+\sin
   \left(f(x)^2\right)  \sin \left(g(x)^2\right)= \cos\left(f(x)^2-g(x)^2 \right)$$ Continue and get the result.
A: The derivatives of $\cos^2x$ and $\sin^2x$ are $\mp\sin2x$, in this order, and we leave them on the side.
Then we get (loosely)
$$(\sin(c^2)\cos(s^2))'\to\cos(c^2)\cos(s^2)--\sin(c^2)\sin(s^2)=\cos(c^2-s^2)$$
(double minus intended). Combining these results, the derivative is indeed
$$-\sin(2x)\cos(\cos(2x)).$$
