I found a really weird function

[Chuckstabler]

Hey all, so I don't even know how I managed to find this, So I'll walk you all through how I've spent my last couple hours.

I've been studying intro calculus, and I recently came across the chain rule. I was screwing around with finding the derivative of various functions like the square-root of 4x^2 + 4. I then took the derivative of square-root 4x^2 + 4 (I think that's the function I took, I screwed up somewhere, but oh well, it doesn't change the function that I found), and found it's derivative using the quotient rule, and after simplifying I was left with

(4/(sqrt(4x^2+4)) - (16x^2/sqrt(4x^2+4)^3/2)

You can find it on at this link: (It's a graphing calculator website), https://www.desmos.com/calculator/l68uv67zay

It looks remarkably similar to a normal distribution, and seems to exhibit some properties of a normal distribution, is it doing this for any particular reason?

 

[Chestermiller]

Why don't you reduce it to a common denominator and also factor out √4 from within the radical? What are you left with then?

Chet

 

[Chuckstabler]

Give me a bit, i'll work on it. Thanks for your comment :D

 

[Svein]

 

[mfb]

Your function can be simplified to$$(x^2+1)^{-\frac{3}{2}}$$
Functions like this don't look completely different from a normal distribution on a linear plot, but they have a completely different shape in the tails.

 

[suremarc]

Your function has second derivative $f''(x)=\frac{-3x}{(x^2+1)^{5/2}}$, which is approximately equal to $-3xf(x)$ when $x$ is small.
The function $g(x)=e^{-\frac{3x^2}{2}}$ has derivative $-3xg(x)$, and furthermore $f'(0)=g(0)=1$; so $g$ can be said to approximate $f'$ near $0$. Coincidentally $g$ describes a normal PDF, so that explains why your function looks like a normal distribution. I would link a graph comparing the two (they are very close!) but I happen to be posting from my phone presently.