Derivative of $(\sin^2(t))^t$ I am trying to solve for the derivative of $G(t) = (\sin^2(t))^t$. My approach is to use the substitutions $u = \sin^2(t)$ and $v = \sin(t)$ yielding the equations
$G(t) = u^t$ and 
$G'(t) = \ln u\cdot u^t\cdot u'$
where $u = v^2$ and $u'=2v\cdot v' = 2\sin t\cos t = \sin 2t$
Using this approach, I get the answer $\ln(\sin^2(t))(\sin^2t)^t\sin 2t$, which is different from the correct answer $2(\sin^2t)^t(\ln\sin t+t\cot t)$. I understand why this answer is correct using logarithmic differentiation.
Could someone explain why my initial (and usual) approach for solving derivatives using chain rule is wrong here? Thanks in advance.
 A: The substitution $u$ depends on $t$.
And you don't have $\frac{d}{dt} u(t)^t = \ln(u(t)) \cdot u^t(t) \cdot u'(t)$, but $$\frac{d}{dt} u(t)^t = t \cdot u(t)^{t - 1} \cdot u'(t) + u(t)^t \log(u(t)).$$ This can be seen through logarithmic differentiation by considering $$\frac{d}{dx} \log(y(x)) = \frac{d}{dx}\log(f(x)^{x}) = \frac{d}{dx} (x \log(f(x)) = \log(f(x)) + \frac{x \cdot f(x)}{f'(x)}.$$
Now $\frac{d}{dx} \log(y(x)) = \frac{y'(x)}{y(x)}$. Solving for $y(x) = f(x)^x$ gives the desired result.
Alternatively, you can write $f(t)^t = e^{t \log(f(t))}$ and then use $\frac{d}{dx} e^{g(x)} = g'(x) e^{g(x)}$, yielding (analogously to above)
$$
\frac{d}{dt} f(t)^t
= \left( \log(f(t)) + \frac{t \cdot f(t)}{f'(t)} \right) f(t)^{t}.
$$

In your case we therefore get
\begin{align}
G'(t)
& = t \cdot (\sin^2(t))^{t - 1} \cdot 2 \sin(t) \cos(t)
+ (\sin^2(t))^{t} \cdot \log(\sin^2(t)) \\
& = 2t \cos(t) \sin^{2t - 1}(t) + \sin^{2t}(t) \cdot 2 \log(\sin(t)) \\
& = 2 \sin^{2t}(t) \left[t \cot(t) + \log(\sin(t)) \right].
\end{align}
