How to calculate $I^{'}\left(0\right)=2\sqrt{2}$, $I\left(x\right)=\intop_{0}^{2x}f\left(x-t\right)dt$,

Question

How to calculate $I^{'}\left(0\right)$, $I\left(x\right)=\intop_{0}^{2x}f\left(x-t\right)dt$, where \begin{equation} f\left(x\right)=\begin{cases} \frac{\ln\left(1+\sqrt{2}x+x^{2}\right)}{x} & x\ne0\\ 0 & x=0 \end{cases} \end{equation}

I did $y=x-t$, then $I\left(x\right)$ becomes $I\left(x\right)=\intop_{-x}^{x}f\left(y\right)dy$. Then I am not sure how to continue.

Then I am not sure how to continue.

The derivative $I'(x)$ at $x=0$ should be calculated by the definition?

Any hint? Thank you very much

Answer 1

Use the Leibnitz theorem to differentiate the integral thus $$I'(x)=f(x-2x)\cdot2-f(x-0)\cdot0=2f(-x)$$

$$ I'\left(x\right)= 2\frac{\ln\left(1-\sqrt{2}x+x^{2}\right)}{-x} , x\ne0 $$

$$I'(0)=\lim_{x \to 0}\frac{2\ln\left(1-\sqrt{2}x+x^{2}\right)}{-x}$$

use LHopital's rule to calculate the limit thus $I'(0)=2\sqrt{2}$