# Differentiating the Dot Product's Evil Twin

Trying2Learn
TL;DR Summary
Title says it all
Hello,

I found this link very useful:
https://www.quora.com/Why-is-cosine-used-in-dot-products-and-sine-used-in-cross-products

I understand all of Anders Kaseorg's discussion except for ONE PONT.

At the very end, he writes: "[the evil twin of the dot product] is not differentiable at parallel vectors."

Could someone explain why? (I can see the issue with the evil twin of the cross, but not the dot)

Homework Helper
Gold Member
Look at his formula for the dot evil twin. It has a square root in it. Square root functions are not differentiable at zero. With a little algebra, it's easy to show that the inside of the square root is zero for parallel vectors, so the function is not differentiable there.

The actual function being used is something like ##\sqrt{|x|}##. If you graph that function on the interval ##[-1,1]## you'll see that it's continuous but not differentiable at ##x=0##. It bounces off the x-axis there.
Also look at the graph of ##|\sin x|## on ##[-1,1]## to see another relevant non-differentiable bounce.

By contrast, the formulas for the usual dot and the usual cross have no square roots, or anything else that can upset differentiability, such as the denominators in the evil twin cross. They pass zero angles (parallel for dot, perp for cross) smoothly.

Trying2Learn
Look at his formula for the dot evil twin. It has a square root in it. Square root functions are not differentiable at zero. With a little algebra, it's easy to show that the inside of the square root is zero for parallel vectors, so the function is not differentiable there.

OH! I see!

THANK YOU!