1
$\begingroup$

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.

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

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.

$\endgroup$
  • 1
    $\begingroup$ thank you, it was a very nice explanation. I understand it now. $\endgroup$ – Electric Mar 16 at 6:31
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $a-b+ab \ne 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.