Answer
$$ G'(t)=(\sin^2t)^t(\ln (\sin^2t)+2t\cot t) .$$
Work Step by Step
Since $ G(t)=(\sin^2t)^t $, applying $\ln $ on both sides, we get
$$\ln G(t)=\ln (\sin^2t)^t=t \ln (\sin^2t)$$
Hence the derivative, using the product rule, is given by
$$ G'/G = \ln (\sin^2t)+\frac{t}{\sin^2t}(\sin^2t)'= \ln (\sin^2t)+\frac{t}{\sin^2t}(2\sin t \cos t) \\
=\ln (\sin^2t)+2t\cot t.$$
Then, we have
$$ G'(t)=G(t)(\ln (\sin^2t)+2t\cot t)=(\sin^2t)^t(\ln (\sin^2t)+2t\cot t) .$$
