0
$\begingroup$

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.

  • 0
    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)?2012-04-20

1 Answers 1

3

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:

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

  • 1
    you're welcome. For future reference, it is not recommended to cancel terms inside parenthesis with terms outside those parenthesis...2012-04-20