## Thomas' Calculus 13th Edition

$\frac{5}{2}$
$g(x)=\sqrt{x}$ $g'(x)=\frac{1}{2\sqrt{x}}$ and $g'(1)=\frac{1}{2}$ $f(u)=u^5+1$ $f'(u)=5u^4$ $f'(g(1))=f'(1)=5$ Therefore $(f o g)'(-1)=f'(g(1))g'(1)=5*\frac{1}{2}=\frac{5}{2}$