<s>[INST] Computing the derivative from the definition Using the limit definition of the derivative which I know is:
$$f'(x)=\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)$$
I am trying to solve this problem 
$$f(x)= \frac{x}{x+2}   $$
How do I go about properly solving this, I seemed to get  

$$\frac{x}{x+2}\   $$
  as my answer again? What are the steps I should follow?

I am trying to find the derivative of $f(x)= \frac{x}{x+2}$ using the definition of the derivative. 
 [/INST] I think that you mean that you want to use the definition of the derivative to find the derivative of the function $$f(x)=\frac{x}{x+2}\;.$$ The definition is that $$f\,'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}h\;.$$ Note the limit: it’s essential.
For your function this is
$$\begin{align*}
\lim_{h\to 0}\frac{\frac{x+h}{(x+h)+2}-\frac{x}{x+2}}h&=\lim_{x\to 0}\left(\frac1h\left(\frac{x+h}{(x+h)+2}-\frac{x}{x+2}\right)\right)\\
&=\lim_{h\to 0}\left(\frac1h\cdot\frac{(x+2)(x+h)-x(x+2+h)}{(x+2)(x+2+h)}\right)\\
&=\lim_{h\to 0}\frac{x^2+2x+hx+2h-x^2-2x-hx}{h(x+2)(x+2+h)}\\
&=\lim_{h\to 0}\frac{2h}{h(x+2)(x+2+h)}\\
&=\lim_{h\to 0}\frac2{(x+2)(x+2+h)}\;,
\end{align*}$$
where I’ve simply put the original numerator $f(x+h)-f(x)$ over a common denominator and simplified. At this point we’ve managed to get rid of the factor of $h$ in the denominator, so we can take the limit: as $h\to 0$, the numerator just sits there at $2$, and the denominator approaches $(x+2)(x+2)$, or $(x+2)^2$. Thus,
$$f\,'(x)=\lim_{x\to 0}\frac2{(x+2)(x+2+h)}=\frac2{(x+2)^2}\;.$$
Added: By the way, if your algebra is good enough, you can notice that $$\frac{x}{x+2}=\frac{(x+2)-2}{x+2}=1-\frac2{x+2}\;.$$ Then 
$$\begin{align*}
f(x+h)-f(x)&=\left(1-\frac2{(x+h)+2}\right)-\left(1-\frac2{x+2}\right)\\
&=\frac2{x+2}-\frac2{x+2+h}\\
&=\frac{2\big(x+2+h-(x+2)\big)}{(x+2)(x+2+h)}\\
&=\frac{2h}{(x+2)(x+2+h)}\;,
\end{align*}$$
and you reach the point of being able to take the limit a bit more easily.
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