Answer
$$
\begin{aligned}
h(x)&=\int_{\sqrt{x}}^{x^{2}} \cos \left(t^{2}\right) d t\\
&=\int_{\sqrt{x}}^{0} \cos \left(t^{2}\right) d t+\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t\\
& =-\int_{0}^{\sqrt{x}} \cos \left(t^{2}\right) d t+\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t
\end{aligned}
$$
the derivative of the function $h(x)$ is given by the following:
$$
\begin{aligned}
h^{\prime}(x)&=-\cos \left((\sqrt{x})^{2}\right) \cdot \frac{d}{d x}(\sqrt{x})+\left[\cos \left(x^{3}\right)^{2}\right] \cdot \frac{d}{d x}\left(x^{3}\right) \\
&=-\frac{1}{2 \sqrt{x}} \cos x+3 x^{2} \cos \left(x^{6}\right)
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
h(x)&=\int_{\sqrt{x}}^{x^{2}} \cos \left(t^{2}\right) d t\\
&=\int_{\sqrt{x}}^{0} \cos \left(t^{2}\right) d t+\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t\\
& =-\int_{0}^{\sqrt{x}} \cos \left(t^{2}\right) d t+\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t ,
\end{aligned}
$$
the derivative of the function $h(x)$ is given by the following:
$$
\begin{aligned}
h^{\prime}(x)&=-\cos \left((\sqrt{x})^{2}\right) \cdot \frac{d}{d x}(\sqrt{x})+\left[\cos \left(x^{3}\right)^{2}\right] \cdot \frac{d}{d x}\left(x^{3}\right) \\
&=-\frac{1}{2 \sqrt{x}} \cos x+3 x^{2} \cos \left(x^{6}\right)
\end{aligned}
$$
