Answer
For $x=1$, $$(fog)'=\frac{5}{2}$$
Work Step by Step
$f(u)=u^5+1$, $u=g(x)=\sqrt x$, $x=1$
Applying the Chain Rule: $$(fog)'=f'(g(x))g'(x)=f'(u)u'$$
We have
$f'(u)=(u^5+1)'=5u^4$
$u'=(\sqrt x)'=(x^{1/2})'=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt x}$
Therefore,
$$(fog)'=5u^4\times\frac{1}{2\sqrt x}=\frac{5(\sqrt x)^4}{2\sqrt x}=\frac{5x^2}{2\sqrt x}=\frac{5}{2}x^{3/2}$$
For $x=1$, $$(fog)'=\frac{5}{2}1^{3/2}=\frac{5}{2}$$
