<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] 
Your last step is wrong, the actual steps will be,

$$=\lim_{h\to 0} \frac{\frac{-\color{red}{(hx^2-h+h^2x)}}{(1+x^2)(x^2+2hx+h^2+1)}}{\color{red}{h}}$$
$$=\lim_{h\to 0} \frac{\frac{-\color{blue}{(x^2-1+hx)}}{(1+x^2)(x^2+2hx+h^2+1)}}{\color{blue}{1}}$$

$$=\frac{1-x^2}{(1+x^2)^2}$$

</s>