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.:) 
 A: To apply the derivative like so, you need to learn a thing called the product rule or quotient rule.  To avoid it, we may proceed as follows:
$$\frac{1-x}{2+x}=\frac{3-(2+x)}{2+x}=\frac3{2+x}-1$$
The derivative of a constant is zero, hence we only need to look at
$$\frac3{2+x}=3(2+x)^{-1}\stackrel{d/dx}\to-3(2+x)^{-2}$$
However, be careful to note that
$$\frac d{dx}(1+2x)^{-1}\ne-(1+2x)^{-2}$$
Since it is not of the form $(c+x)^n$.  Power rule works specifically, for now, for derivatives of the following form:
$$\frac d{dx}(c+x)^n=n(c+x)^{n-1}$$
Where $x$ is positive with a coefficient of one.
A: You can try this
$\frac{d}{dx}f(x) = \frac{d}{dx}(-1+\frac{3}{2+x}) = 3\frac{d}{dx}(2+x)^{-1} = \frac{-3}{(x+2)^2}$
A: 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.
A: 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.
