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.:) 

You can try this
$\frac{d}{dx}f(x) = \frac{d}{dx}(-1+\frac{3}{2+x}) = 3\frac{d}{dx}(2+x)^{-1} = \frac{-3}{(x+2)^2}$
