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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.

  • 0
    Please fix the title.2019-05-26

2 Answers 2

6

$y = \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = \dfrac{e^x}{e^x} \times \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = \dfrac{e^{2x} - 1}{e^{2x} + 1} = \dfrac{e^{2x} + 1 - 2}{e^{2x} + 1} = 1 - \dfrac2{e^{2x}+1}$ \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}

  • 2
    The correct form is `\operatorname{sech}`.2012-11-23
3

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

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

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