## Calculus (3rd Edition)

$$G'(t)=(\sin^2t)^t(\ln (\sin^2t)+2t\cot t) .$$
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 $$G'(t)=G(t)(\ln (\sin^2t)+2t\cot t)=(\sin^2t)^t(\ln (\sin^2t)+2t\cot t) .$$