Answer
a) $\dfrac{d}{dx}[ \tan^{-1} x(\sinh x) +C] =sech x$
b) $\dfrac{d}{dx}[ \sin^{-1} x(\tanh x) +C] =sech x$
Work Step by Step
(a) Since, $\cosh^2 x-\sinh^2 x=1$
$\dfrac{d}{dx}[ \tan^{-1} x(\sinh x) +C]=\dfrac{1}{(\sinh x)^2 +1} \cosh x=\dfrac{\cosh x}{\sinh^2 x +1}$
or, $ =sech x$
(b) Since, $1- \tanh^2 x =sech^2 x$
$\dfrac{d}{dx}[ \sin^{-1} x(\tanh x) +C]=\dfrac{1}{\sqrt {1-\tanh^2 x}} sech^2 x=\dfrac{sech^2 x}{\sqrt{sech^2 x}} $
or, $=sech x$
Hence, a) $\dfrac{d}{dx}[ \tan^{-1} x(\sinh x) +C] =sech x$
b) $\dfrac{d}{dx}[ \sin^{-1} x(\tanh x) +C] =sech x$
