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.

  • 2
    $\begingroup$ "Derivative of a function", not "derivative of an equation". $\endgroup$
    – Jean Marie
    Commented Mar 16, 2019 at 6:10

3 Answers 3


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 :


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


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.

  • 1
    $\begingroup$ thank you, it was a very nice explanation. I understand it now. $\endgroup$
    – Electric
    Commented Mar 16, 2019 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 $a-b+ab \ne 0$


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