🧵 Matrix calculus
Anonymous at Sun, 16 Jun 2024 22:45:08 UTC No. 16238531
Why is it so hard to find resources on matrix calculus? the only actually useful resource I could find was the derivatives chapter of the matrix cookbook, yet I see matrix calulus used in many places (eg. in optimisation problems where the derivative is used).
It seems like there is some secret resource everyone practices to get good at it but I can't find it. Is there a proper book on this with problems and solutions? And no, it's not the same as single variable or multivariable calculus because we're differentiating with respect to a matrix
Anonymous at Sun, 16 Jun 2024 22:46:11 UTC No. 16238536
>>16238531
You have to be over 18 to post here.
Anonymous at Sun, 16 Jun 2024 22:50:06 UTC No. 16238541
>>16238536
matrix calculus is not studied by under 18s, retard
at best they would learn single variable calculus
Anonymous at Sun, 16 Jun 2024 22:53:20 UTC No. 16238551
>>16238541
>He relies on school for his education.
Yikes. Double yikes. Infinite yikes.
Anonymous at Sun, 16 Jun 2024 22:55:04 UTC No. 16238556
>>16238551
go back to plebbit you dribbling retard
you don't even know multivariable calculus neverminds anything relevant to the question
Anonymous at Sun, 16 Jun 2024 22:58:55 UTC No. 16238563
>>16238556
>Thinks multivariable calculus is some big thing.
>>>/mlp/
Anonymous at Mon, 17 Jun 2024 00:47:11 UTC No. 16238703
>>16238531
You do a derivative with a matrix. My linalg class spent a few hours on this, even code monkeys know basic matrix calculus
>>16238541
In first world countries people learn multivariable calculus in hs
Anonymous at Mon, 17 Jun 2024 00:54:39 UTC No. 16238711
>>16238531
https://www.youtube.com/playlist?li
All it takes is a fucking search you faggot.
Anonymous at Mon, 17 Jun 2024 01:02:44 UTC No. 16238716
>>16238531
It is just like you would think.
You just need to keep things in the same order since matrix multiplication isn't commutative.
Product rule: (fg)' = f'g+fg'
Power rule: (f^n)' = f'*f^(n-1) + f*f'*f^(n-2) + ... + f^(n-1)*f'