Differentiate Fraction: Finding the Derivative of 2/(x+1)

[max0005]

Homework Statement

Find the first derivate of $$\frac{2}{x+1}$$.

Formula: $$\frac{f(x+h)-f(x)}{h}$$

Homework Equations

Formula: $$\frac{f(x+h)-f(x)}{h}$$

The Attempt at a Solution

$$\frac{\frac{2}{x+h+1}-\frac{2}{x+1}}{h}$$

$$h(\frac{2}{x+h+1}-\frac{2}{x+1})$$

$$h(\frac{2(x+1)-2(x+h+1)}{(x+1)(x+h+1)}$$

$$h*\frac{-2h}{x^2+xh+2x+h+1}$$

...How do I go on from here? I should get rid of all h in the denominator, but don't have any clue on how to do that... :(

 

[eumyang]

First off, this is in the wrong forum. Should be in the Calculus and Beyond sub forum.

Second, the formula is wrong. The limit part is missing.

Third, this step:
$$h(\frac{2}{x+h+1}-\frac{2}{x+1})$$
is wrong. Should be
$$\frac{1}{h}(\frac{2}{x+h+1}-\frac{2}{x+1})$$

 

[max0005]

First off, this is in the wrong forum. Should be in the Calculus and Beyond sub forum.

Second, the formula is wrong. The limit part is missing.

Third, this step:
$$h(\frac{2}{x+h+1}-\frac{2}{x+1})$$
is wrong. Should be
$$\frac{1}{h}(\frac{2}{x+h+1}-\frac{2}{x+1})$$

Then I'd get

$$\frac{1}{h}*\frac{-2h}{x^2+xh+2x+h+1}$$

But I don't understand how to continue..

 

[The Chaz]

Cancel a factor of h from top and bottom. Then let h -> 0 and you'll get
$$\frac{-2}{x^2+2x+1} = \frac{-2}{(x + 1)^2}$$

Later, when you learn "shortcuts" (i.e rules of differentiation that will be proven), this second form will look a lot more familiar.

 

[eumyang]

Then I'd get

$$\lim_{h \rightarrow 0} \frac{1}{h}*\frac{-2h}{x^2+xh+2x+h+1}$$

But I don't understand how to continue..

(Fixed the above for you.)
Can you see that something cancels?

EDIT: Never mind, The Chaz beat me to it...

 

[max0005]

Ok, got it, thanks! :D