[GRE 1268, #32a] Let \(f(x)= \displaystyle \int_{2x}^{x^2}t^tdt\). Find \(f'(2)\).

Solution Whenever you see an integral where one or both the boundaries contains a variable, think of the Fundamental Theorem of Calculus (part I): \begin{equation}\label{eq} \frac{d}{dx}\int_a^x g(t)dt =g(x),\end{equation} where \(a\) is any fixed constant. In fact, in this problem we need a more general version of it, that is easily derived as follows. Let's call the integral on the left of equation (\ref{eq}) \(G(x)\): \[G(x)=\int_a^x g(t)dt, \hspace{2ex} .\] Then the FTC (part I) says that \(G'(x)= g(x)\). If \(h(x)\) is any differentiable function with appropriate domain, so that the composition \(G(h(x))\) is defined, we can use the chain rule to find \begin{equation} \label{eq2} \frac{d}{dx} \int_a^{h(x)} g(t) dt=\left(G\circ h\right)'(x) = G'(h(x)) h'(x) =g(h(x)) h'(x).\end{equation} We can easily generalize this to an integral where both boundaries are non-constant. Given two functions \(h_1(x)\) and \(h_2(x)\), take any \(a\) such that all integrals below are defined. Then we can write \[\int_{h_1(x)}^{h_2(x)} g(t)dt = \int_{h_1(x)}^a g(t)dt + \int_a^{h_2(x)}g(t)dt = \int_a^{h_2(x)}g(t)dt - \int_a^{h_1(x)} g(t)dt .\] Using the formula from equation (\ref{eq2}), we find the more general form of the Fundamental Theorem of Calculus that we need for this problem: \begin{equation}\label{eq3} \frac{d}{dx} \int_{h_1(x)}^{h_2(x)}g(t)dt = g(h_2(x))h_2'(x)-g(h_1(x))h_1'(x).\end{equation} Applying (\ref{eq3}) to \(g(t) = t^t\), \(h_1(x)=2x\), \(h_2(x)=x^2\), we find \[ g(2x)=(2x)^{2x}, \hspace{2ex} g(x^2) = (x^2)^{x^2}, \hspace{2ex} h_1'(x)=2,\hspace{2ex} h_2'(x)=2x,\] and so the derivative of the given \(f(x)\) is \[f'(x)=2x(x^2)^{x^2}-2 (2x)^{2x}\] and then \[f'(2)=4(4)^4-2(4)^4=2(4)^4=512.\]