Answer
For $x=0$, $$(fog)'=0$$
Work Step by Step
$f(u)=\frac{2u}{u^2+1}$, $u=g(x)=10x^2+x+1$, $x=0$
Applying the Chain Rule: $$(fog)'=f'(g(x))g'(x)=f'(u)u'$$
We have
$$f'(u)=\Big(\frac{2u}{u^2+1}\Big)'=\frac{(2u)'(u^2+1)-2u(u^2+1)'}{(u^2+1)^2}$$ $$f'(u)=\frac{2(u^2+1)-2u\times2u}{(u^2+1)^2}=\frac{2(u^2+1)-4u^2}{(u^2+1)^2}$$ $$f'(u)=\frac{2(-u^2+1)}{(u^2+1)^2}$$
$$u'=(10x^2+x+1)'=20x+1$$
For $x=0$, we have $$u'(0)=20\times0+1=1$$
Also, for $x=0$, $u(0)=10\times0^2+0+1=1$
Therefore, $$f'(1)=\frac{2(-1^2+1)}{(1^2+1)^2}=\frac{2\times0}{2^2}=0$$
That means,
$$(fog)'=0\times1=0$$
