Question:

<b>Inverse Function Derivative</b><br> Since $g$ is the inverse of $f$: $g(f(x)) = x$ for all $x$.<br> Differentiating: $g'(f(x))\cdot f'(x) = 1$, so $g'(f(x)) = \dfrac{1}{f'(x)}$ whenever $f'(x)\neq 0$.<br> Check each option:<br> (A) $g'(f(0)) = 1$? Only if $f'(0)=1$. $f'(x)=4(2x-3x^2)^3(2-6x)-\sin x$, $f'(0)=4\cdot 0\cdot 2-0=0$. So $g'(f(0))$ is undefined — (A) false.<br> (B) $g'(f(0)) = 1/f'(0)$: true by the inverse function theorem (stated as an identity) — (B) true.<br> (C) $g'(f(1)) = 1/f'(1)$: $f(1)=(2-3)^4+\cos 1=1+\cos 1$; $f'(1)=4(-1)^3(-4)-\sin 1=16-\sin 1\neq 0$ — (C) true.<br> (D) $g'(f(x))\cdot f'(x)=1$: This is the direct derivative of $g(f(x))=x$ — always true — (D) true.<br> <b>Key concept:</b> $(g\circ f)'(x) = g'(f(x))\cdot f'(x) = 1$ is the fundamental inverse-function identity.<br> <b>Trap:</b> Option (A) claims $g'(f(0))=1$ but that requires $f'(0)=1$; check before assuming.