1. The problem is to find the derivative of the function $f(\sin 5x)$.\n\n2. We use the chain rule for derivatives, which states that if $f(g(x))$ is a composite function, then its derivative is $f'(g(x)) \cdot g'(x)$.\n\n3. Here, $f$ is a function of $\sin 5x$, so let $u = \sin 5x$. Then $f(u)$ is the outer function and $u = \sin 5x$ is the inner function.\n\n4. The derivative of $f(u)$ with respect to $u$ is $f'(u)$. The derivative of $u = \sin 5x$ with respect to $x$ is $\cos 5x \cdot 5$ by the chain rule.\n\n5. Therefore, the derivative of $f(\sin 5x)$ with respect to $x$ is:\n$$\frac{d}{dx} f(\sin 5x) = f'(\sin 5x) \cdot \frac{d}{dx} (\sin 5x) = f'(\sin 5x) \cdot 5 \cos 5x.$$\n\n6. This is the final answer: $$f'(\sin 5x) \cdot 5 \cos 5x.$$