Derivative of a integral with limits as functions of other variable

Question

I'm being asked to calculate, explicitly the following derivative:

$$ \frac{d}{d x}\left[\int_{e^{-x}}^{e^{x}} \sqrt{1+(\ln t)^{2}} d t\right] $$

How can I do that?

I thought about using the fundamental theorem of calculus and chain rule but I'm not sure about this approach. I think that the integral itself can not be done analyticalally, so I can not integrate it first to do the derivative latter.

Answer 1

Let $f(x)$ be a continuous function (where it is needed).

Then $F'(x)=f(x)$, where $F(x)=\int_a^xf(t)dt$ (for some appropiate $a$). This is FTC.

Now, $$G(x)=\int_{a(x)}^{b(x)}f(t)dt=\int_a^{b(x)} f(t)dt- \int_a^{a(x)}f(t)dt =F(b(x))-F(a(x))$$ (or simply, use Barrow). So, applying Chain rule, and taking into account that $F'=f$: $$G'(x)=F'(b(x))b'(x)-F'(a(x))a'(x)=f(b(x))b'(x)-f(a(x))a'(x).$$ Now, apply THAT formula to your situation.

Answer 2

$$\frac{d}{dx}\int_{e^{-x}}^{e^x} \sqrt{1+(\ln t)^2} dt= e^x \sqrt{1+x^2}+e^{-x} \sqrt{1+x^2}=\sqrt{1+x^2}(e^x+e^{-x})=2\cosh x \sqrt{1+x^2}$$

Note that: $$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt= a'(x) f(a(x))-b'(x) f(b(x))$$