The derivative of $\tanh x$ I'm trying to calculate the derivative of $\displaystyle\tanh h = \frac{e^h-e^{-h}}{e^h+e^{-h}}$. Could someone verify if I got it right or not, if I forgot something etc. Here goes my try:
$$\frac{d}{dh}\left( \frac{e^h - e^{-h}}{e^h + e^{-h}}\right) = \frac{d}{dh}\left( (e^h - e^{-h})\cdot(e^h + e^{-h})^{-1}\right) $$ $$= \frac{d}{dh}\left( e^h -e^{-h}\right)\cdot(e^h + e^{-h})^{-1} + (e^h -e^{-h})\cdot\frac{d}{dh}\left( (e^h + e^{-h})^{-1}\right) $$$$ = (e^h + e^{-h})\cdot(e^h + e^{-h})^{-1} - (e^h-e^{-h})\cdot\frac{e^h-e^{-h}}{(e^h+e^{-h})^2} $$$$= 1-\frac{(e^h-e^{-h})^2}{(e^h+e^{-h})^2} = 1-\frac{(e^{2h}-2 + e^{-2h})}{e^{2h} + 2+e^{-2h}} =  1+\frac{-e^{2h}+2 - e^{-2h}}{e^{2h} + 2+e^{-2h}}$$
Thnx for any help! :) 
 A: When faced with multiplications and divisions in a differentiation, use logarithmic differentiation:
$$\begin{align} 
\\ \tanh x &= \frac{e^x-e^{-x}}{e^x+e^{-x}}
\\ \ln (\tanh x) &= \ln (\frac{e^x-e^{-x}}{e^x+e^{-x}})
\\ \ln (\tanh x) &= \ln ({e^x-e^{-x}}) - \ln ({e^x+e^{-x}})
\\ \frac{\mathrm d}{\mathrm dx} ( \ln (\tanh x) &= \frac{\mathrm d}{\mathrm dx} \ln ({e^x-e^{-x}}) - \frac{\mathrm d}{\mathrm dx} \ln ({e^x+e^{-x}}) 
\\ \frac{\mathrm d ( \ln (\tanh x)}{\mathrm d( \tanh x)} \frac{\mathrm d ( \tanh x)}{\mathrm d( x)}  &= \frac{\mathrm d}{\mathrm dx} \ln ({e^x-e^{-x}}) - \frac{\mathrm d}{\mathrm dx} \ln ({e^x+e^{-x}}) 
\\ \frac{1}{\tanh x} \frac{\operatorname d ( \tanh x)}{\operatorname d( x)}  &= \frac{\operatorname d \ln ({e^x-e^{-x}})}{ \operatorname  d ({e^x-e^{-x}})} \frac{\operatorname d ({e^x-e^{-x}})}{ \operatorname  d x} - \frac{\operatorname d \ln (e^x+e^{-x}) }{\operatorname d (e^x+e^{-x})} \frac{\operatorname d (e^x+e^{-x}) }{\operatorname d x} 
\\ \frac{1}{\tanh x} \frac{\operatorname d ( \tanh x)}{\operatorname d( x)}  &= \frac{ 1}{   ({e^x-e^{-x}})} ({e^x+e^{-x}}) - \frac{1 }{(e^x+e^{-x})} (e^x-e^{-x})  
\\ \frac{1}{\tanh x} \frac{\operatorname d ( \tanh x)}{\operatorname d( x)}  &= \frac{( e^x+e^{-x})(e^x+e^{-x}) }{   ({e^x-e^{-x}})(e^x+e^{-x})} - \frac{(e^x-e^{-x})(e^x-e^{-x}) }{(e^x+e^{-x})(e^x-e^{-x})}   
\\ \frac{1}{\tanh x} \frac{\operatorname d ( \tanh x)}{\operatorname d( x)}  &= \frac{( e^x+e^{-x})^2 }{   ({e^x-e^{-x}})(e^x+e^{-x})} - \frac{(e^x-e^{-x})^2 }{(e^x+e^{-x})(e^x-e^{-x})}
\\ \frac{1}{\tanh x} \frac{\operatorname d ( \tanh x)}{\operatorname d( x)}  &= \frac{4}{   ({e^x-e^{-x}})(e^x+e^{-x})} 
\\  \frac{\operatorname d ( \tanh x)}{\operatorname d(x)}  &= \frac{4 \tanh x}{   ({e^x-e^{-x}})(e^x+e^{-x})} \square
\end{align} $$
A: There are two ways of tackling the problem. One way is to expand $\tanh x$:
$$
\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}=
\frac{e^x-e^{-x}}{e^x+e^{-x}}\frac{e^x}{e^x}=
\frac{e^{2x}-1}{e^{2x}+1}
$$
and then using the quotient rule. Tedious, but easy.
The second way is to remember that
$$
\tanh x=\frac{\sinh x}{\cosh x}
$$
and again using the quotient rule, but taking into account that the derivatives of $\sinh$ and $\cosh$ are…
