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Suppose I have the following hyperbolic tangent function: $$\ f(x)=\frac{(1-e^{-2x})}{(1+e^{-2x})}$$

What will be the first derivative of this function?

Mosk
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maddy
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  • Just a minor nitpick: that isn't an equation, since there's no equality. It's an expression. – Zubin Mukerjee Apr 20 '15 at 00:09
  • First of all, welcome to MSE. When posting a question, please also show attempts of solving it for yourself. Note that formally, we can only derive functions, not equations/expressions. Now have you tried the quotient rule for derivatives? ($(f/g)'=\frac {gf'-fg'}{g^2}$). – Uncountable Apr 20 '15 at 00:13
  • You can find the answer here, but you could also derive it by hand as user robjohn does quite nicely. Wikipedia also mentions that the hyperbolic tangent is the solution to the system $$f' = 1-f^2$$ $$f(0)=0$$ – Zubin Mukerjee Apr 20 '15 at 00:16

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The derivative is a bit simpler to compute if we write it as $$ \begin{align} \frac{1-e^{-2x}}{1+e^{-2x}} &=\frac{e^{2x}-1}{e^{2x}+1}\\ &=1-\frac2{e^{2x}+1} \end{align} $$ Taking the derivative using the chain-rule gives $$ \begin{align} \frac2{(e^{2x}+1)^2}2e^{2x} &=\frac4{(e^x+e^{-x})^2}\\[6pt] &=\mathrm{sech}^2(x) \end{align} $$

robjohn
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