Finding The Derivative Using $\frac{d}{dx}x^n=nx^{n-1}$ So I am learning how to differentiate now,
and I came across this problem
$$f(x)=\frac{1-x}{2+x}$$ 
We are wanted to find $f'(x)$.
When I use $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
I find that $f'(x)=\frac{-3}{(2+x)^2}$ but, when I try to find $f'(x)$ the easy way.
i.e $\frac{d}{dx}x^=nx^{n-1}$. I cannot do it for some reason. Is it not possible to use that derivatives property when dealing with quotients? I know we can use it polynomial addition, subtraction and multiplication but I am struggling with quotients. Can someone please explain what it is I am not seeing?
My Attempt:
$$\frac{d}{dx}\frac{1-x}{2+x}=\frac{d}{dx}(1-x)(2+x)^{-1}=(1)(-1)(2+x)^{-2}$$
Which is obviously wrong so can someone please break this down for me.:) 

There might be a misconception lurking here. Maybe not. In either case, I hope this helps.
Yes, by all means, the limit definition of the derivative is
$$ f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
Do you want a derivative of some arbitrary (well behaved) function? Say $x^2-3x$ or $\sin x$ or $e^x$? Then, sure, the above limit will give you the derivative. At least, that's true in theory. You've probably demonstrated this for polynomials in $x$, but the limit is more involved for other functions and can require a good deal of analysis.
Anyway, you next cite the "easy" method for the derivative,
$$\frac{d}{dx}x^n=nx^{n-1}$$
This isn't the derivative of any function, though. It's specifically the rule for power laws. That is, it only works for functions of the form $x^n$. It will not work, in general, for rational functions. As the others have answered, though, you might be able to use the power law with rational functions after polynomial long division.
