Answer
$$
f^{\prime}\left(\frac{\pi}{2}\right)=-1
$$
Work Step by Step
$$
f(x)=\int_{0}^{g(x)} \frac{1}{\sqrt{1+t^{3}}} d t, \text { where } g(x)=\int_{0}^{\cos x}\left[1+\sin \left(t^{2}\right)\right] d t
$$
Differentiating both sides of this equation, using $FTC1$ and the Chain Rule (twice) we have
$$
\begin{aligned}
f^{\prime}(x)&=\frac{1}{\sqrt{1+[g(x)]^{3}}} g^{\prime}(x)\\
&=\frac{1}{\sqrt{1+[g(x)]^{3}}} \frac{d}{dx} \left[\int_{0}^{\cos x}\left[1+\sin \left(t^{2}\right)\right] d t \right]\\
&=\frac{1}{\sqrt{1+[g(x)]^{3}}}\left[1+\sin \left(\cos ^{2} x\right)\right](-\sin x) .
\end{aligned}$$
Now
$$ g\left(\frac{\pi}{2}\right)=\int_{0}^{\cos (\frac{\pi}{2})}\left[1+\sin \left(t^{2}\right)\right] d t=\int_{0}^{0}\left[1+\sin \left(t^{2}\right)\right] d t=0,
$$
so
$$
\begin{aligned}
f^{\prime}\left(\frac{\pi}{2}\right)&=\frac{1}{\sqrt{1+[g(\frac{\pi}{2})]^{3}}}\left[1+\sin \left(\cos ^{2} (\frac{\pi}{2})\right)\right](-\sin( \frac{\pi}{2})) \\
&=\frac{1}{\sqrt{1+0}}(1+\sin 0)(-1) \\
&=1 \cdot 1 \cdot(-1)\\
&=-1
\end{aligned}
$$
