Answer
For $x=1$, $$(fog)'=-\frac{\pi}{4}$$
Work Step by Step
$f(u)=\cot\frac{\pi u}{10}$, $u=g(x)=5\sqrt x$, $x=1$
Applying the Chain Rule: $$(fog)'=f'(g(x))g'(x)=f'(u)u'$$
We have
$$f'(u)=\Big(\cot\frac{\pi u}{10}\Big)'=-\csc^2\frac{\pi u}{10}\Big(\frac{\pi u}{10}\Big)'=-\frac{\pi}{10}\csc^2\frac{\pi u}{10}$$
$$u'=(5\sqrt x)'=(5x^{1/2})'=\frac{5}{2}x^{-1/2}=\frac{5}{2\sqrt x}$$
Therefore,
$$(fog)'=\Big(-\frac{\pi}{10}\csc^2\frac{\pi u}{10}\Big)\frac{5}{2\sqrt x}=-\frac{\pi}{10}\Big(\csc^2\frac{\pi(5\sqrt x)}{10}\Big)\frac{5}{2\sqrt x}$$
$$=-\frac{5\pi}{20\sqrt x}\Big(\csc^2\frac{\pi\sqrt x}{2}\Big)=-\frac{\pi}{4\sqrt x}\csc^2\frac{\pi\sqrt x}{2}$$
For $x=1$, $$(fog)'=-\frac{\pi}{4\times\sqrt1}\csc^2\frac{\pi\sqrt1}{2}=-\frac{\pi}{4}\csc^2\frac{\pi}{2}$$
$$=-\frac{\pi}{4}\times\frac{1}{\sin^2\frac{\pi}{2}}=-\frac{\pi}{4}\times\frac{1}{1^2}=-\frac{\pi}{4}$$
