How can i take ds/dx and ds/dy of a function numerically

I tried to take derivatives of s=-x^3+3*x*y^2 numericaly

ds/dx=3*y^2-3*x^2

ds/dy=6*x*y

I tried to gradient of s to find ds/dx and ds/dy but results are far from analtical results. My code is below

clc;clear;
[x y]=meshgrid(0:5,0:5)
s=-x.^3+3.*x.*y.^2
[dsx,dsy]=gradient(s)

dsy should refers ds/dy

0 3 6 9 12 15

0 6 12 18 24 30

0 12 24 36 48 60

0 18 36 54 72 90

0 24 48 72 96 120

0 27 54 81 108 135

ds/dy =6xy analtically

0 0 0 0 0 0

0 6 12 18 24 30

0 12 24 36 48 60

0 18 36 54 72 90

0 24 48 72 96 120

0 30 60 90 120 150

How can i take derivatives numerically in gradient way. where is my fault.

NOTE: It is the simplified example to find the derivatieves numerically. My real project is very complicated and all values in matrices form. So I do not intrested in analtical solutions. It should be solved in numerical. Why gradient results are far away from analtical results how can i solve this numerically.

[Ameer Hamza]

The numerical gradient can not be exactly equal to the analytical gradient, except in very simple cases. For any nonlinear function, there will be a huge difference between both. There is no way around except using a very small step size, which will decrease the difference beween analytical and numerical gradient. For example:

If step size is 1

[x, y] = meshgrid(0:5,0:5);
s = -x.^3+3.*x.*y.^2;
[~, dsy] = gradient(s, 1, 1);
err = mean((dsy - 6.*x.*y).^2, 'all') % mean squared error

Result

>> err
err =
   27.5000

If step size is 0.1

[x, y] = meshgrid(0:0.1:5,0:0.1:5);
s = -x.^3+3.*x.*y.^2;
[~, dsy] = gradient(s, 0.1, 0.1);
err = mean((dsy - 6.*x.*y).^2, 'all') % mean squared error

Result

>> err
err =
    0.0297

[esat gulhan]

Oh yes, the error is step size. thanks

[Ameer Hamza]

I am glad that this solved the issue for you. I have added it as an answer too.

Answer 1

[Ameer Hamza]

The numerical gradient can not be exactly equal to the analytical gradient, except in very simple cases. For any nonlinear function, there will be a huge difference between both. There is no way around except using a very small step size, which will decrease the difference beween analytical and numerical gradient. For example:

If step size is 1

[x, y] = meshgrid(0:5,0:5);
s = -x.^3+3.*x.*y.^2;
[~, dsy] = gradient(s, 1, 1);
err = mean((dsy - 6.*x.*y).^2, 'all') % mean squared error

Result

>> err
err =
   27.5000

If step size is 0.1

[x, y] = meshgrid(0:0.1:5,0:0.1:5);
s = -x.^3+3.*x.*y.^2;
[~, dsy] = gradient(s, 0.1, 0.1);
err = mean((dsy - 6.*x.*y).^2, 'all') % mean squared error

Result

>> err
err =
    0.0297

Answer 2

[KSSV]

ds/dx=-3*x^2+3*y^2
ds/dy = 6*x*y 

[esat gulhan]

I know analtical solution. How can i solve it numerically. I take gradient but it gives wrong result