Derivative of function and simplification

I'm attempting to find the derivative of the following function:

$$\frac{e^x - e^{-x}}{e^x + e^{-x}}$$

where I would use the quotient rule to find its derivative. In the end, I would obtain:

$$\frac{(e^x + e^{-x})^2 -(e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$

However, I checked with the calculator online and I found out that the equation I current have above can be further simplified to:

$$\text{ Rewrite / Simplify }$$ $$=1-\frac{(e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$ $$\text{ Simplify }$$ $$\frac{4e^{2x}}{(e^{2x} +1)^2}$$

I'm having issues understanding the simplification step above.

My biggest question is, in the second image above, why couldn't I have $$(e^x + e^{-x})^2$$ cancel out $$(e^x + e^{-x})^2$$ at the denominator, which would leave me with just $$-(e^x - e^{-x})^2$$? Could anyone explain the simplification step to me as I am very much confused.

P.S. pardon me but I couldn't understand the syntax on writing equations here so I would opt to use screenshot instead.

• "Derivative of a function", not "derivative of an equation". – Jean Marie Mar 16 at 6:10

I think your issue is that the computation in the last "cartridge" should be decomposed into $$2$$ steps :

First step :

The first expression can be written, taking a common denominator :

$$\dfrac{(e^x+e^{-x})^2}{(e^x+e^{-x})^2}+\dfrac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=\dfrac{(e^x+e^{-x})^2+(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=\dfrac{4}{(e^x+e^{-x})^2}$$

Second step : Factor $$e^{-x}$$ in the denominator :

$$(e^x+e^{-x})^2=(e^{-x}(e^{2x}-1))^2=e^{-2x}(e^{2x}-1)^2$$

We have obtained a $$(e^{2x}-1)^2$$ as a factor in the denominator.

Getting rid of $$e^{-2x}$$ in the denominator by transforming it into $$e^{2x}$$ in the numerator, explains the final result with $$4e^{2x}$$ in the numerator.

• thank you, it was a very nice explanation. I understand it now. – Electric Mar 16 at 6:31

Hint: Use that $$a^2-b^2=(a-b)(a+b)$$

So it's not even about derivative, as your question is that why:

$$\frac{a - b}{a} \ne -b, a>0$$

Well, they are just not equal, because in general $$a-b+ab \ne 0$$