# Find the Derivative of $$f(x) = \int_{0}^{x^{2}} \frac{\sin(z)}{z} \mathrm{d}z$$

How to find the derivative of

f(x) = \int\limits_{0}^{x^{2}} \frac{\sin(z)}{z} \mathrm{d}z

was a question posed at Mathematics Stack Exchange. There is an obvious solution posted there, but here I present an alternative solution. While it is less efficient than the answer at MSE it is interesting.

Using the series definition of the sine function we have

\sin(z) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} x^{2n-1}

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

\int\limits_{0}^{x^{2}} z^{2n-2} \mathrm{d}z = \frac{1}{2n-1} (x^{2})^{2n-1}

and now we have

f(x) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!(2n-1)} (x^{2})^{2n-1}

Let

g(x) = (x^{2})^{2n-1}

and

\frac{\mathrm{d}g}{\mathrm{d}x} = \frac{2}{x}(2n-1)(x^{2})^{2n-1}

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}