Answer
$0$
Work Step by Step
$(fog)'(x)$ = $f'(g(x))g'(x)$
$g(x)$ = $10x^{2}+x+1$
$g(0)$ = $10(0)^{2}+0+1$ = $1$
$g'(x)$ = $20x+1$
$g'(0)$ = $20(0)+1$ = $1$
$f(u)$ = $\frac{2u}{u^{y2}+1}$
$f'(u)$ = $\frac{d}{dx}$[$\frac{2u}{u^{2}+1}$]
$f'(u)$ = $\frac{({u^{2}+1})\frac{d}{dx}(2u)-(2u)\frac{d}{dx}({u^{2}+1})}{({u^{2}+1})^{2}}$
$f'(u)$ = $\frac{({u^{2}+1})(2)-(2u)({2u})}{({u^{2}+1})^{2}}$ = $\frac{2u^{2}+2-4u^{2}}{({u^{2}+1})^{2}}$ = $\frac{-2u^{2}}{({u^{2}+1})^{2}}$
$f'(1)$ = $f'(g(0))$ = $\frac{-2u^{2}+2}{({u^{2}+1})^{2}}$ = $\frac{-2(1)^{2}+2}{({(1)^{2}+1})^{2}}$ = $0$
$(fog)'(0)$ = $f'(g(0))g'(0)$ = $(0)$$(1)$ = $0$