<s>[INST] Finding derivative of $\frac{x}{x^2+1}$ using only the definition of derivative I think the title is quite self-explanatory. I'm only allowed to use the definition of a derivative to differentiate the above function. Sorry for the formatting though.

Let $f(x) = \frac{x}{x^2+1}$
$$
\begin{align}
f'(x)&= \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0} \frac {\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}\\
&= \lim_{h\to 0} \frac{\frac{x+h}{x^2+2hx+h^2+1}-\frac{x}{x^2+1}}{h}\\
&= \lim_{h\to 0} \frac{\frac{(x+h)(x^2+1)-x(x^2+2hx+h^2+1)}{(x^2+1)(x^2+2hx+h^2+1)}}{h}\\
&= \lim_{h\to 0} \frac{\frac{hx^2-2hx-h^2+h}{(x^2+1)(x^2+2hx+h^2+1)}}{h}
\end{align}
$$

I'm currently stuck with simplify the fraction so that I can finally find the derivative. I'd really appreciate some advice on how to proceed with the problem. 
 [/INST] \begin{eqnarray}
{1 \over h } ({x+h \over 1 + (x+h)^2} - {x \over 1+x^2}) &=&
{1 \over h } { (x+h)(1+x^2) - (1+(x+h)^2)x\over (x+x^2) (1 + (x+h)^2} \\
&=& {1 \over h } { h -h x^2 - h^2 x\over (x+x^2) (1 + (x+h)^2} \\
&=& { 1 - x^2 - h x\over (x+x^2) (1 + (x+h)^2}
\end{eqnarray}
The limit follows by taking $ h \to 0$.
</s>