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here is the function I'm trying to diferentiate at 1 (it wouldn't fit in the title) $$F(x)= \begin{cases} {\frac{\pi x^2}{4}} &\quad\text{if }1\ge x\ge0, \\ \newline \sqrt{x^2-1}+ \frac{x^2}{2}(\frac{\pi}{2}-2\arccos(\frac1x))&\quad\text{if }\sqrt{2}\ge x>1 \\ \end{cases}$$

The derivatives of each of the branches works out to the following: $$f(x)= \begin{cases} {\frac{\pi x}{2}} &\quad\text{if }1\ge x\ge0, \\ \newline \frac{x}{\sqrt{x^{2} - 1}} + \left(\frac{\pi}{2} - \arccos\left(\frac{1}{x}\right)\right) x - \frac{\pi}{2 \sqrt{1 - \frac{1}{x^{2}}}}&\quad\text{if }\sqrt{2}\ge x>1\\ \end{cases}$$

What's left to show is that the derivative by definition at 1 approaching from the right evaluates to $\pi/2$.

What I've tried: The simpler methods of working with limits such as L'Hopital at various stages, distributing and regrouping, or attempting to rewrite some of the fractions yielded 0 progress. Plotting the function in Desmos shows something that looks diferentiable. graph of F

Motivation: Taking a unit square an centering the bottom left corner at (0,0), I want to find the average distance from the origin to a random point (a simpler version of the original problem I set out to solve, finding the average distance between 2 random points in the square).

I'm aproaching this using probability distributions, $F(x)$ being the probability that a point is at most distance $x$ from the origin (this was derived geometrically and it's very possible a simpler formulation exists). Then to find the expected value I wanted to find $f(x)$ through diferentiation and calculating $$\mathbb{E(x)}=\int_{-\infty}^\infty xf(x)dx$$ Any help on this problem would be very apreciated.

RobPratt
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var
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  • Care that for the first branch, the derivative should be $\pi,x$. – Davide Aug 30 '24 at 18:50
  • @Davide my bad, the derivative is fine but the F branch should've been $\frac{\pi x^2}{4}$ as in one quarter of the circle with radius x. edited – var Aug 30 '24 at 19:04
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    I suspect your $- \dfrac{\pi}{2 \sqrt{1 - \frac{1}{x^{2}}}}$ should be $- \dfrac{1}{ \sqrt{1 - \frac{1}{x^{2}}}}$ which is equal to $-\dfrac{x}{\sqrt{x^{2} - 1}}$, leaving $\left(\frac{\pi}{2} - \arccos\left(\frac{1}{x}\right)\right) x$ which is $1$ when $x=1$ – Henry Aug 30 '24 at 19:09
  • The original motivating question---the expected distance between two random points in a square---may be found on MSE here. – Semiclassical Aug 30 '24 at 19:19

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