Derivative of an integral function with exponential upper limit

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Given the integral function $F(x)=\int_1^{e^{2x}}\ln t\,dt,$ compute $F'(x)$, its stationary points and any inflection points of $F$.

Solution

Derivative. We apply the fundamental theorem of calculus together with the chain rule. The upper limit is $u(x)=e^{2x}$, with $u'(x)=2e^{2x}$, and the integrand is $\ln t$. Hence $F'(x)=\ln\big(u(x)\big)\cdot u'(x)=\ln\big(e^{2x}\big)\cdot 2e^{2x}=2x\cdot 2e^{2x}=4x\,e^{2x}.$

Stationary points. Since $e^{2x}>0$ for every $x$, $F'(x)=4x\,e^{2x}=0 \iff x=0.$

Nature of the stationary point. Differentiating again: $F''(x)=4e^{2x}+4x\cdot 2e^{2x}=4e^{2x}(1+2x).$ At $x=0$ we get $F''(0)=4>0$, so $x=0$ is a minimum.

Inflection points. Candidate inflections satisfy $F''(x)=0$: $4e^{2x}(1+2x)=0 \iff 1+2x=0 \iff x=-\tfrac12.$ Around $x=-\tfrac12$ the factor $(1+2x)$ changes sign, so it is an inflection point.

$\boxed{F'(x)=4x\,e^{2x},\qquad x_{staz}=0\ (\text{minimum}),\qquad x_{fl}=-\tfrac12}$