<s>[INST] Derivative of $f(x)=\int^{x^2}_0 \frac{\sin(t)}{t}dt$ 
Let $f(x)=\int^{x^2}_0 \frac{\sin(t)}{t}dt$. Find $f'(x)$.


Is that integral undefined/nonexistant, or just impossible to integrate?
In this case does $f'(x)$ exist and can be solved normally?
Is it correct that $f'(x)=\frac{\sin(x^2)}{x^2}2x$?
 [/INST] The answer provided by @Clarinetist is the most efficient way to solve this problem. Here is another way.
Using the series definition of the sine function we have
\begin{equation}
\sin(z) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} x^{2n-1}
\end{equation}
and
\begin{align}
f(x) &= \int\limits_{0}^{x^{2}} \frac{\sin(z)}{z} \mathrm{d}z \\
&= \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} \int\limits_{0}^{x^{2}} z^{2n-2} \mathrm{d}z
\end{align}
The integral on the right hand side above is
\begin{equation}
\int\limits_{0}^{x^{2}} z^{2n-2} \mathrm{d}z = \frac{1}{2n-1} (x^{2})^{2n-1}
\end{equation}
and now we have
\begin{equation}
f(x) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!(2n-1)} (x^{2})^{2n-1}
\end{equation}
Let
\begin{equation}
g(x) = (x^{2})^{2n-1}
\end{equation}
then
\begin{equation}
\frac{\mathrm{d}g}{\mathrm{d}x} = \frac{2}{x}(2n-1)(x^{2})^{2n-1}
\end{equation}
Now we have
\begin{align}
\frac{\mathrm{d}f}{\mathrm{d}x} &= \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!(2n-1)} \frac{\mathrm{d}g}{\mathrm{d}x} \\
&= \frac{2}{x} \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} (x^{2})^{2n-1} \\
&= \frac{2}{x} \sin(x^{2})
\end{align}
</s>