Answer
$f'(\pi/2)=-1$
Work Step by Step
Find $g'(x)$ using the Chain Rule and the Fundamental Theorem of Calculus in Part 1:
$g'(x)=\frac{d}{dx}\int_0^{\cos x}[1+\sin(t^2)]dt$
$g'(x)=\frac{d}{d(\cos x)}\int_0^{\cos x}[1+\sin(t^2)]dt\cdot \frac{d(\cos x)}{dx}$
$g'(x)=[1+\sin(\cos^2 x)]\cdot (-\sin x)$
$g'(x)=-[1+\sin(\cos^2 x)]\sin x$
Find $f'(x)$ using the Chain Rule and the Fundamental Theorem of Calculus in Part 1:
$f'(x)=\frac{d}{dx}\int_0^{g(x)}\frac{1}{\sqrt{1+t^3}}dt$
$f'(x)=\frac{d}{d(g(x))}\int_0^{g(x)}\frac{1}{\sqrt{1+t^3}}dt\cdot \frac{d(g(x))}{dx}$
$f'(x)=\frac{1}{1+[g(x)]^3}\cdot \left(-[1+\sin(\cos^2 x)]\sin x\right)$
Find $g(\pi/2)$:
$g(\pi/2)=\int_0^{\cos (\pi/2)}[1+\sin((\pi/2)^2)]dt=\int_0^0[1+\sin(\pi^2/4)]dt=0$
Find $f'(\pi/2)$:
$f'(\pi/2)=\frac{1}{1+[g(\pi/2)]^3}\cdot \left(-[1+\sin(\cos^2 (\pi/2))]\sin \pi/2\right)$
$f'(\pi/2)=\frac{1}{1+0^3}\cdot \left(-[1+\sin 0]\cdot 1\right)$
$f'(\pi/2)=\frac{1}{1}\cdot \left(-[1+0]\cdot 1\right)$
$f'(\pi/2)=1\cdot -1$
$f'(\pi/2)=-1$