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 will need to apply the quotient rule. Let the whole function be denoted $h(x)=\frac{f(x)}{g(x)}$, the numerator in your fraction be $f(x)$ and the denominator be $g(x)$. Then the derivative of your function can be found by the following forumla: 
$$h'(x)=\frac{g'(x)f(x)-f'(x)g(x)}{(g(x)^2}$$. Note that you can compute the derivative of $f(x)$ and $g(x)$ individually using the method you have described above, and can simply plug the results into the equation.
