# How do you show that the derivative of arctan(tanhx) = sech (2x)?

Mar 17, 2015

The Chain Rule implies that

$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \arctan \left(\setminus \tanh \left(x\right)\right)\right) = \setminus \frac{1}{1 + \setminus {\tanh}^{2} \left(x\right)} \setminus \cdot \setminus \frac{d}{\mathrm{dx}} \left(\setminus \tanh \left(x\right)\right) .$

Since $\setminus \frac{d}{\mathrm{dx}} \left(\setminus \tanh \left(x\right)\right) = \setminus {\sech}^{2} \left(x\right)$, this becomes

$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \arctan \left(\setminus \tanh \left(x\right)\right)\right) = \setminus \frac{\setminus {\sech}^{2} \left(x\right)}{1 + \setminus {\tanh}^{2} \left(x\right)} .$

To see that this equals $\setminus \sech \left(2 x\right)$, you could note that

$\setminus \sech \left(2 x\right) = \setminus \frac{1}{\setminus \cosh \left(2 x\right)} = \setminus \frac{2}{{e}^{2 x} + {e}^{- 2 x}}$

and

$\setminus \frac{\setminus {\sech}^{2} \left(x\right)}{1 + \setminus {\tanh}^{2} \left(x\right)} = \setminus \frac{\setminus \frac{1}{{\cosh}^{2} \left(x\right)}}{1 + \setminus \frac{\setminus {\sinh}^{2} \left(x\right)}{\setminus {\cosh}^{2} \left(x\right)}} = \setminus \frac{1}{\setminus {\cosh}^{2} \left(x\right) + \setminus {\sinh}^{2} \left(x\right)}$

But this is the same as

$\setminus \frac{1}{{\left(\setminus \frac{{e}^{x} + {e}^{- x}}{2}\right)}^{2} + {\left(\setminus \frac{{e}^{x} - {e}^{- x}}{2}\right)}^{2}} = \setminus \frac{4}{{e}^{2 x} + 2 + {e}^{- 2 x} + {e}^{2 x} - 2 + {e}^{- 2 x}} = \setminus \frac{2}{{e}^{2 x} + {e}^{- 2 x}} .$