Simplify the following exponential expression Simplify: $${e^{-2x}-1\over e^x+1}$$
The question I am trying to answer is actually concerned with finding the derivative of this expression. However, simplifying this expression before differentiating makes that task trivial. The problem is that I cannot find a way to simplify this expression into the equivalent expression, ${1-e^x\over e^{2x}}$, or $-e^{-2x}(e^x-1)$. 
I have a feeling I will feel quite stupid when I see what I am missing but any help would be greatly appreciated. 
 A: $${e^{-2x}-1\over e^x+1}$$
$$=e^{-2x}\left(\frac{1-e^{2x}}{e^x+1}\right)$$
$$=e^{-2x}\left(\frac{(1-e^{x})(1+e^{x})}{e^x+1}\right)$$
$$=e^{-2x}(1-e^x)$$
A: Here is, perhaps, a bit of intuition:  the difficulty with taking the derivative of the given expression is that it is a quotient, which means that if you don't do anything to it, you are going to have to attack it with the chain and power rules (or the quotient rule), which can make things quite messy.  To make our lives easier, we would like to somehow find a factor of $e^x + 1$ upstairs in order to kill off the denominator.
If you examine the term that is upstairs, it looks like a difference of squares.  So we can factor it as
$$ e^{-2x} - 1 = (e^{-x})^2 - 1^2 = (e^{-x} + 1)(e^{-x} -1).$$
This is close to what we want, but still not there.  If we could turn the $e^{-x} + 1$ into an $e^{x}+1$ somehow, we would be in business.  We can do this by multiplying by 1 in a clever way:
$$ e^{-x} + 1 = 1 \cdot (e^{-x} +1) = \frac{e^x}{e^x}(e^{-x} + 1) = \frac{1}{e^x}(1+e^x) = e^{-x}(1+e^x).$$
Substituting all of this back into the original expression, we get
$$ \frac{e^{-2x} - 1}{e^x + 1}
 = \frac{(e^{-x} - 1)(e^{-x} + 1)}{e^x + 1}
 = \frac{(e^{-x} -1) e^{-x} (1+e^x)}{e^x + 1}
 = e^{-x}(e^{-x}-1).
$$
Personally, I would further simplify this to
$$ e^{-2x} - e^{-x}, $$
since I would much rather take the derivative of a sum than a product, but you could also multiply by $e^{-x}$ upstairs and downstairs to get the result that others have given.
