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Claim: $${d\over dx}\tanh^{-1}x = {1 \over 1-x^2}$$

My work. Let $y = \tanh^{-1}x$. Then $$\tanh y =x \Rightarrow {d \tanh y \over dy}\cdot {dy\over dx} = 1$$ Since $\tanh y = \dfrac{e^y - e^{-y}}{e^y + e^{-y}}$ then $${d \tanh y \over dy} = d {(e^y - e^{-y})(e^y + e^{-y})^{-1} \over dy}=1 - (e^y - e^{-y})(e^y + e^{-y})^{-2}(e^y - e^{-y})= 0$$ which means $0 \cdot {dy\over dx} = 1$

I think I did something wrong but I cannot find which part it is.

Please, give me some hints or advice.

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  • $\begingroup$ $\tanh^{-1}(x)=\dfrac12\ln\left(\dfrac{1+x}{1-x}\right)=-i\tan^{-1}(ix)$ $\endgroup$ – Bumblebee Nov 4 '17 at 5:35
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The problem is $$ (e^y - e^{-y})(e^y + e^{-y})^{-2}(e^y - e^{-y}) \neq 1 $$ An alternative avoiding exponentials is $$ \frac{d \tanh(y)}{dy}=\frac{d \sinh(y)/\cosh(y)}{dy}=\frac{\cosh^2(y)-\sinh^2(y)}{\cosh^2(y)}=1-\tanh^2(y)=1-x^2 $$

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You approach is correct but $${d \tanh y \over dy} = {d \over dy} \left(\frac{e^y - e^{-y}}{e^y + e^{-y}}\right)=\frac{(e^y + e^{-y})(e^y + e^{-y})-(e^y - e^{-y})(e^y - e^{-y})}{(e^y + e^{-y})^2}\\=\frac{4}{(e^y + e^{-y})^2}=\frac{1}{\cosh^2 y}.$$ Now note that $$\tanh^2 y=\frac{\sinh^2 y}{\cosh^2 y}=\frac{\cosh^2 y-1}{\cosh^2 y}=1-\frac{1}{\cosh^2 y}$$ where we used the identity $\cosh^2 y-\sinh^2 y=1$. Can you take it from here?

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  • $\begingroup$ I found my problem. Thx $\endgroup$ – Beverlie Jun 29 '17 at 9:48
  • $\begingroup$ How could one enlarge the size of letter like you in case of fractional expression? $\endgroup$ – Beverlie Jun 29 '17 at 9:52
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    $\begingroup$ @Beverlie Use \dfrac{}{} command tex.stackexchange.com/questions/135389/… $\endgroup$ – Robert Z Jun 29 '17 at 9:57
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Notice that \begin{align} (e^y-e^{-y})(e^y+e^{-y})^2(e^y-e^{-y}) &= \left( \frac{e^y+e^{-y}}{e^y-e^{-y}}\right)^2 \\ &= \frac{1}{\tanh^2y} \end{align}

Then proceed from there, you have the correct method, and have tried to show from a greater detail than relying on standard derivatives and for that you should be applauded.

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I allow myself to post this answer that I like and I don't think this has already been mentionned:

We know that $$\tanh(\tanh^{-1}(x)) = x$$ Taking the derivative on both sides : $$ \tanh'(\tanh^{-1}(x)) \cdot \tanh^{-1}(x)' = 1 \Leftrightarrow \tanh^{-1}(x)' = \frac{1}{\tanh'(\tanh^{-1}(x)) } $$ Knowing that $\tanh'(x) = 1-\tanh(x)^2$ we get : $$ \tanh^{-1}(x)' = \frac{1}{\tanh'(\tanh^{-1}(x)) } = \frac{1}{1-\tanh^2(\tanh^{-1}(x))} = \frac{1}{1-x^2} $$

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