Finding The Derivative Using $\frac{d}{dx}x^n=nx^{n-1}$

Question

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.:)

Answer 1

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.

Answer 2

You can try this

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

Answer 3

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.

Answer 4

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.