Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am curious as how to solve this

I have been trying here and I know the answer is $\dfrac{4}{{(e^x+e^{-x}})^2}$ I have this derivative $y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$.

So here is how we do it. Now I understand I have to use the product rule but I have no idea how we got to the solution of $$\dfrac{4}{{(e^x+e^{-x}})^2}$$

So I know we use the product rule but im not sure how they got to $4$ for the top part of the fraction.

share|cite|improve this question
Please fix the title. – TMM Nov 22 '12 at 20:07
up vote 6 down vote accepted

$$y = \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = \dfrac{e^x}{e^x} \times \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = \dfrac{e^{2x} - 1}{e^{2x} + 1} = \dfrac{e^{2x} + 1 - 2}{e^{2x} + 1} = 1 - \dfrac2{e^{2x}+1}$$ \begin{align} \dfrac{dy}{dx} & = - 2 \dfrac{d}{dx} \left(\dfrac1{1+e^{2x}}\right)\\ & = - 2 \dfrac{d}{d e^{2x}} \left(\dfrac1{1+e^{2x}}\right) \dfrac{d(e^{2x})}{dx}\\ & = -2 \left(\dfrac{-1}{(1+e^{2x})^2}\right) \times 2 e^{2x}\\ & = \dfrac{4e^{2x}}{(1+e^{2x})^2} = \dfrac{4}{\dfrac{(1+e^{2x})^2}{e^{2x}}} = \dfrac{4}{\left(\dfrac{1+e^{2x}}{e^{x}} \right)^2}\\ & = \dfrac4{(e^{-x} + e^x)^2} \end{align}

share|cite|improve this answer
Note the function to be differentiated is $\tanh x$ and the result $\mbox{sech}^2 \; x.$ I do not seem to have a code for sech in Latex, so I used mbox. Probably this MathJax has something. – Will Jagy Nov 22 '12 at 20:14
The correct form is \operatorname{sech}. – kahen Nov 23 '12 at 3:53

A little knowledge of hyperbolic functions can help a lot to simplify stuff here:

$$\frac{e^x−e^{−x}}{e^x+e^{−x}}=\frac{\frac{e^x−e^{−x}}{2}}{\frac{e^x+e^{−x}}{2}}=\frac{\sinh x}{\cosh x}=:\tanh x\Longrightarrow$$

$$\Longrightarrow (\tanh x)'=\frac{1}{\cosh^2x}=\frac{4}{(e^x+e^{-x})^2}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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