Take the derivative of a SYMBOLIC Matrix with respect to a Vector

Question

N/A 2023-3-21

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1932500-take-the-derivative-of-a-symbolic-matrix-with-respect-to-a-vector

评论： Bjorn Gustavsson 2023-3-22

采纳的回答： Bjorn Gustavsson

在 MATLAB Online 中打开

Hi community,

I am trying to compute the derivative of a matrix

with respect to a vector

. Both have symbolic components. I cannot use the naive 'for-loop' implementation because the matrix is quite large and, more importantly, the

and in general is quite complex (many trigonometric functions). I was wondering if there is a faster 'vectorized' implementation to perform this computation.

% PSEUDOCODE of WHAT I HAVE 
% M = matrix function of (x) size: [nxn]
% x vector size: [nx1]
n = length(x)
for i = 1:n
    for j = 1:n 
        for k = 1:n
        M_ij = M(i,j)   
        dM_dx(i,j,k) = diff(M_ij, x(k))
        end
    end
end

I would like to know if there is a method to perform this computation without for loops, like the following:

% PSEUDOCODE of WHAT I WISH
dM_dx = diff(M, x)

PS: I already tried to look at: Differentiate symbolic expression or function - MATLAB diff (mathworks.com) but doesn't seem to work since it gives me the following error:

'Components of second argument explicitly present in first argument.'

Thank you very much for the help!

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Answer 1

Bjorn Gustavsson 2023-3-21

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1932500-take-the-derivative-of-a-symbolic-matrix-with-respect-to-a-vector#answer_1197610

编辑：Bjorn Gustavsson 2023-3-21

在 MATLAB Online 中打开

You might get close enough with jacobian to get the full output:

syms x y real
% simple 2 x 2 matrix:
M = [sin(x+y) cos(x*y);-cos(x/(x^2+y^2)) tan(x)];
% Should return this:
% M =
% 
% [         sin(x + y), cos(x*y)]
% [-cos(x/(x^2 + y^2)),   tan(x)]
% 
% This will turn the 2x2 matrix M into a 4x1 column-array, which jacobian
% can handle:
J = jacobian(M(:),[x,y])
% should return 
% J =
%  
% [                                                cos(x + y),                                cos(x + y)]
% [sin(x/(x^2 + y^2))*(1/(x^2 + y^2) - (2*x^2)/(x^2 + y^2)^2), -(2*x*y*sin(x/(x^2 + y^2)))/(x^2 + y^2)^2]
% [                                               -y*sin(x*y),                               -x*sin(x*y)]
% [                                              tan(x)^2 + 1,                                         0]

Here you get all 4 x 2 derivatives, you will have to re-sort the rows into the ordering you expect. For this toy-model-example I get:

J3D = reshape(J,[2,2,2]);

to produce something like what you describe. In your case this might be a rather big array of the size n1 x n2 x n_vars - if I understood your problem right.

HTH

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N/A 2023-3-22

Thank you very much. It solves my problem :)

Bjorn Gustavsson 2023-3-22

在 MATLAB Online 中打开

Good that it solved your problem!

It just dawned on me that it is possible to use the (:)-operation for the assignment to avoid the hassle with reshaping the jacobian:

syms x y real
% Symbolic matrix easier to differentiate manually:
M = [x*y, x^2*y x^3*y;x*y^2, x^2*y^2, x^3*y^2];
% define the size of the Jacobian-like array:
syms JM [2 3 2]
% doing it in one swoop!(?):
JM(:) = jacobian(M(:),[x,y])