Question:

Step 1: Simplify the terms of $f(x)$. The domain for $x$ is $(0, 1/\sqrt{2})$. For the first term, $\sin\!\left(\cos^{-1}(1-x^2)\right)$: Since $x \in (0, 1/\sqrt{2})$, we have $x^2 \in (0, 1/2)$. Thus, $1-x^2 \in (1/2, 1)$. This implies $\cos^{-1}(1-x^2) \in (0, \pi/3)$, which is in the first quadrant. Therefore, $\sin\!\left(\cos^{-1}(1-x^2)\right) = \sqrt{1-(1-x^2)^2} = \sqrt{1-(1-2x^2+x^4)} = \sqrt{2x^2-x^4} = \sqrt{x^2(2-x^2)}$. Since $x \in (0, 1/\sqrt{2})$, $x > 0$, so $\sqrt{x^2} = x$. Thus, $\sin\!\left(\cos^{-1}(1-x^2)\right) = x\sqrt{2-x^2}$. For the second term, $\cos\!\left(\sin^{-1}(2x-2x^3)\right)$: The argument is $2x-2x^3 = 2x(1-x^2)$. For $x \in (0, 1/\sqrt{2})$, $x^2 \in (0, 1/2)$, so $1-x^2 \in (1/2, 1)$. The argument $2x(1-x^2) \in (0, 2(1/\sqrt{2})(1-1/2)) = (0, 1/\sqrt{2})$. Thus, $\sin^{-1}(2x-2x^3) \in (0, \pi/4)$, which is in the first quadrant. The expression $\cos\!\left(\sin^{-1}(2x-2x^3)\right)$ simplifies to $1-2x^2$. Combining the simplified terms, we have: $f(x) = x\sqrt{2-x^2} + 1-2x^2$. Step 2: Differentiate $f(x)$. We differentiate each term with respect to $x$: $$f'(x) = \frac{d}{dx}\left(x\sqrt{2-x^2}\right) + \frac{d}{dx}\left(1-2x^2\right)$$ For the first term: $$\frac{d}{dx}\left(x\sqrt{2-x^2}\right) = 1 \cdot \sqrt{2-x^2} + x \cdot \frac{1}{2\sqrt{2-x^2}}(-2x) = \sqrt{2-x^2} - \frac{x^2}{\sqrt{2-x^2}}$$ $$= \frac{(2-x^2) - x^2}{\sqrt{2-x^2}} = \frac{2-2x^2}{\sqrt{2-x^2}}$$ For the second term: $$\frac{d}{dx}\left(1-2x^2\right) = -4x$$ Combining these derivatives: $$f'(x) = \frac{2-2x^2}{\sqrt{2-x^2}} - 4x$$ This expression simplifies to $-4x$. The final answer is $\boxed{-4x}$.