Problem: Let $f$ is a function defined on $[0,10]$ by
\[
f(x) = \int _{x^2}^{x^4} e^{\sqrt{t} } \mathrm{d} t.
\]
Show that $f$ is differentiable on $(0,10)$ and find the derivative function.
Consider a function $$g(s) \coloneqq \int _0^s e^{\sqrt{t} }\mathrm{d} t.$$ This function is differentiable and use the fundamental theorem of calculus to find the derivative.
Express $f$ in terms of $g$.
Solution: We need to show that the function
\begin{equation}\label{eq:19May2025-1}
f(x) = \int _{x^2}^{x^4} e^{\sqrt{t} } \mathrm{d} t
\end{equation}
is differentiable on $(0,10)$. Note that the function $t \mapsto e^{\sqrt{t} }$ is a continuous function on $\mathbb{R} $ and hence it is integrable on $\mathbb{R} $ or in particular on $[0,10]$. By the Fundamental Theorem of Calculus, the function
\begin{equation}\label{eq:19May2025-2}
g(s) \coloneqq \int _0^s e^{\sqrt{t} }\mathrm{d} t
\end{equation}
is differentiable on $(0,10)$ with the derivative $g'(s) = e^{\sqrt{s} }$. Now,
\begin{align*}
f(x) = \int_{x^2}^{x^4} e^{\sqrt{t} }\mathrm{d} t & = \int_{0}^{x^4} e^{\sqrt{t} } \,\mathrm{d}t - \int_{0}^{x^2} e^{\sqrt{t}} \,\mathrm{d}t \\
& = g\left( x^4 \right) - g\left( x^2 \right) .
\end{align*}
Since $g$ is differentiable, by the chain rule $f$ is also differentiable on $(0, 10)$. Furthermore,
\begin{align*}
f'(x) & = g'\left( x^4 \right) \cdot (x^4)' - g'\left( x^2 \right) \cdot (x^2)' \\
& = e^{\sqrt{x^4} } \cdot 4x^3 - e^{\sqrt{x^2} } \cdot 2x \\
& = 4x^3 e^{x^2} - 2xe^x .
\end{align*}