Answer
$\dfrac{t^2+1}{2t^2}$
Work Step by Step
Given: $G(t)=\dfrac{d}{dt}\sinh {(\ln t)}$
Now, $G'(t)=\dfrac{d}{dt}\sinh {(\ln t)}$
On applying the chain rule, we have
$G'(t)=(\dfrac{d\sinh {(\ln t)}}{d\ln t})(\frac{d\ln t}{dt})$
or, $=\cosh {(\ln t)} \times \dfrac{1}{t}$
or, $=\dfrac{\cosh {(\ln t)}}{t}$
Since, $\cosh x=\dfrac{e^x+e^{-x}}{2}$
$G'(t)=\dfrac{\dfrac{e^{\ln t}+e^{-\ln t}}{2}}{t}=\dfrac{t+\frac{1}{t}}{2t}$
or, $=\dfrac{\frac{t^2+1}{t}}{2t}$
or, $=\dfrac{t^2+1}{2t^2}$
