Answer
$tanh^3 v$
Work Step by Step
Since, $\dfrac{d}{dx} (\cosh x)=\sinh x$; $\dfrac{d}{dx} (\tanh x)=sech^2 x$ and $\tanh^2 x+ sech^2 x=1$
$\dfrac{dy}{dv} =\dfrac{d}{dv} [\ln \cosh v-\dfrac{1}{2} \tanh^2 v)]$
or, $=\dfrac{1}{(\cosh v)}(\sinh v)-\dfrac{1}{2}(2) \tanh v (sech^2 v)$
or, $=\tanh v (1-sech^2 v)$
Thus, $\dfrac{dy}{dv}=tanh^3 v$
