# Derivative of $\frac{x}{1+x^2}$ using the limit definition.

I just started learning Calculus on my own and understand where $$\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$$ comes from, but I'm having trouble with this one; I think my Algebra skills are letting me down.

I start with $$\lim_{h \rightarrow 0} \left(\left(\frac{x+h}{1+(x+h)^2} - \frac{x}{(1+x)^2}\right) \frac{1}{h} \right)$$ but then get lost expanding everything. I don't see how to end up with $$h$$ as a factor in the numerator so that I can get rid of the denominator.

• Do you want to find the derivative of $\frac{1}{1+x^2}$ or $\frac{x}{1+x^2}$? – Paul May 19 '14 at 0:01
• @Paul, thanks for pointing that out, I edited it. – jeremy radcliff May 19 '14 at 0:02
• Note that your difference quotient is not written quite correctly. You can edit the question to put the $\cdot \frac{1}{h}$ outside of the parentheses. The difference $f(x + h) - f(x)$ is all divided by $h$. – Sammy Black May 19 '14 at 0:26

Hint. \eqalign{\frac{f(x+h)-f(x)}{h} &=\Bigl(\frac{x+h}{1+(x+h)^2}-\frac{x}{1+x^2}\Bigr)\frac{1}{h}\cr &=\frac{(x+h)(1+x^2)-x(1+(x+h)^2)}{h(1+x^2)(1+(x+h)^2)}\cr &=\frac{1+x^2-2x^2-hx}{(1+x^2)(1+(x+h)^2)}\cr} and now it is easy to take $h\to0$.
• Is $f(x)$ supposed to be $x/(1+x)^2$ or $x/(1+x^2)$? The title of your question looks like it's the former, so $f(x+h)$ would be $(x+h)/(1+x+h)^2$. – David May 19 '14 at 0:38