Answer
$$G'(t)=\frac{t^2+1}{2t^2}$$
Work Step by Step
$G'(t)=\frac{d}{dt}\sinh {(\ln t)}$
Using the chain rule:
$G'(t)=\frac{d\sinh {(\ln t)}}{d\ln t} \times \frac{d\ln t}{dt}$
$=\cosh {(\ln t)} \times \frac{1}{t}$
$=\frac{\cosh {(\ln t)}}{t}$
Recall that:
$$\cosh x=\frac{e^x+e^{-x}}{2}$$
Thus:
$G'(t)=\frac{\frac{e^{\ln t}+e^{-\ln t}}{2}}{t}$
$=\frac{t+\frac{1}{t}}{2t}$
$=\frac{\frac{t^2+1}{t}}{2t}$
$=\frac{t^2+1}{2t^2}$
