# Why does $-e^x$ disappear?

I have this math problem I am trying to understand, I am supposed to derivate this:

$$f(x) = \frac{e^x}{1+x}$$

And I know the answer and how to work it:

$$f'(x) = \frac{e^x (1+x) - e^x \cdot 1}{(1 + x)^2}$$

The problem I have is that I can't explain what happens next ->

http://yeyfiles.net/547701918/math.png

If I should do it, I'd think $e^x - e^x = 0$

can someone explain to me in an easy way what's happening? :)

Thanks.

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Parentheses, please. Presumably your first line should be e^x/(1+x), but it is not clear to me what the line about f' is supposed to say. And why the (1+0)? –  Ross Millikan Apr 20 '12 at 14:33

You are correct, $e^x-e^x$ is zero, but the way you're showing that cancellation in the image you linked is not clear.

$$f'(x)=\frac{e^x(1+x)-e^x}{(1+x)^2}$$

distribute:

$$f'(x)=\frac{e^x+xe^x-e^x}{(1+x)^2}$$

then cancel:

$$f'(x)=\frac{xe^x}{(1+x)^2}$$

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oh thank you, that was an excellent answer and it does make me feel stupid, which is great :) –  cubsink Apr 20 '12 at 14:37
you're welcome. For future reference, it is not recommended to cancel terms inside parenthesis with terms outside those parenthesis... –  Andrew Parker Apr 20 '12 at 14:38