Question

Show that \(\frac{d}{{dx}}\arctan \left( {\tanh x} \right) = {\mathop{\rm sech}\nolimits} 2x\).

Short answer

It is proved that \(\frac{d}{{dx}}\arctan \left( {\tanh x} \right) = {\mathop{\rm sech}\nolimits} 2x\).

Step-by-step solution

Write the formula of inverse hyperbolic functions

\(\frac{d}{{dx}}\left( {{{\tanh }^{ - 1}}x} \right) = \frac{1}{{{x^2} - 1}}\)

Find the derivative of the function

Differentiate the function\(\arctan \left( {\tanh x} \right)\)with respect to \(x\) and use \({\cosh ^2}x + {\sinh ^2}x = \cosh 2x\) to simplify.

\(\begin{aligned}\frac{d}{{dx}}\arctan \left( {\tanh x} \right) & = \frac{1}{{1 + {{\left( {\tanh x} \right)}^2}}}\frac{d}{{dx}}\left( {\tanh x} \right)\\ & = \frac{{{{{\mathop{\rm sech}\nolimits} }^2}x}}{{1 + {{\tanh }^2}x}}\\ & = \frac{{\frac{1}{{{{\cosh }^2}x}}}}{{1 + \frac{{{{\sinh }^2}x}}{{{{\cosh }^2}x}}}}\\ & = \frac{1}{{{{\cosh }^2}x + {{\sinh }^2}x}}\\ & = \frac{1}{{\cosh 2x}}\\ & = {\mathop{\rm sech}\nolimits} 2x\end{aligned}\)

Hence, it is proved that \(\frac{d}{{dx}}\arctan \left( {\tanh x} \right) = {\mathop{\rm sech}\nolimits} 2x\).