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 $f'(x)=\frac{\sin(x^2)}{x^2}2x$?
 A: Notice that if we call $$g(u) = \int^{u}_0 \frac{sin(t)}{t}dt$$ then we know that $g'(u)=\frac{\sin u}{u}$
But $f(x)=g(x^2)$. Then you apply the chain rule to get your result.

Is that integral undefined/nonexistant

The integral is perfectly defined. It just cannot be expressed in closed form, in terms of elementary functions. But that does not matter at all.
A: By the fundamental theorem of calculus, if $F$ is an antiderivative of $f$, then
$$\int_{\ell(x)}^{u(x)}f(t)\text{ d}t = F(u(x)) - F(\ell(x))\text{.}$$
It follows that, after applying the chain rule,
$$\dfrac{\text{d}}{\text{d}x}\int_{\ell(x)}^{u(x)}f(t)\text{ d}t = F^{\prime}(u(x))u^{\prime}(x) - F^{\prime}(\ell(x))\ell^{\prime}(x)\text{.}$$ 
but since $F$ is an antiderivative of $f$, it follows that $F^{\prime} = f$, hence
$$\dfrac{\text{d}}{\text{d}x}\int_{\ell(x)}^{u(x)}f(t)\text{ d}t = f(u(x))u^{\prime}(x) - f(\ell(x))\ell^{\prime}(x)\text{.}$$ 
This rule obviously requires many conditions (notice, in this case, $f(\ell(x))$ is undefined), so we use a slight modification here: observe that if $F$ is an antiderivative of $\dfrac{\sin(x)}{x}$, $F(0)$ is obviously a constant; then,
$$\dfrac{\text{d}}{\text{d}x}\int_{0}^{x^2}\underbrace{\dfrac{\sin(t)}{t}}_{f(t)}\text{ d}t = \underbrace{\dfrac{\sin(x^2)}{x^2}}_{f(x^2)}\underbrace{(2x)}_{\text{deriv. of }x^2} - \dfrac{\text{d}}{\text{d}x}[F(0)] = \dfrac{2\sin(x^2)}{x}\text{.}$$
A: For a simpler and fundamental question, do you know what is $g'(x)$ when 
$$
g(x)=\int^{x}_0 \frac{\sin(t)}{t}\ dt?
$$
In general, do you know what is $g'(x)$
$$
g(x)=\int_0^xh(t)\ dt?\tag{*}
$$
Once you know what is $g'(x)$ for (*), can you find by chain rule what is the derivative of $f(x)=g(x^2)$?
A: 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}
