To take the partial derivative of a function using matlab

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Pranjal Pathak
Pranjal Pathak 2013-2-11
编辑： Sergio E. Obando 2024-6-15
Here is a particular code. Can anyone please help me in taking the analytical (partial) derivative of the function 'F' along X (i.e., w.r.t. X) along Y (i.e., w.r.t. Y) and along the diagonal (i.e., w.r.t. X plus w.r.t. Y) using matlab command.
[X, Y]=meshgrid(-1:2/511:+1, -1:2/511:+1);
F=sqrt(3).*(2.*(X.^2+Y.^2)-1);
Thanking You!

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采纳的回答

Walter Roberson
Walter Roberson 2013-2-11
If you have the symbolic toolkit
syms X Y
F=sqrt(3).*(2.*(X.^2+Y.^2)-1);
diff(F,X)
diff(F,Y)
diff(F,X,Y)
  5 个评论
显示 3更早的评论
Olivar Luis Eduardo
Olivar Luis Eduardo 2023-4-25
I think that numerical calculation is being requested, not the symbolic one. In other words, the surface is a matrix
Sergio E. Obando
Sergio E. Obando 2024-6-15
Ran in:
For clarification, the numerical and symbolic calculations are fairly similar code wise. Walter has provided the symbolic gradients (as requested), while Youssef has provided the numerical ones. One correction is that dX = matlabFunction(diff(F,x)) is only a function of x, which is why it generates an error when calling dX(x,y).
To go in a bit more detail on Walter's suggested solution:
clear
% Define Mesh
[ X,Y] = meshgrid(-1:2/10:1,-1:2/10:1); % Using 2/10 as spacing
% Analytical Solution (i.e. symbolic var and fnc)
syms F(x,y)
F(x,y) = sqrt(3).*(2.*(x.^2+y.^2)-1);
fsurf(F) % You don't have to pass X or Y
% Analytical Gradients and Hessian
dFdx = diff(F,x)
dFdx(x, y) = 
dFdy = diff(F,y)
dFdy(x, y) = 
gradF = gradient(F)
gradF(x, y) = 
HessF = hessian(F)
HessF(x, y) = 
% or Second Order Derivatives
dFdxdy = diff(F,x,y)
dFdxdy(x, y) = 
0
% Evaluate gradients on a given range
dFdX = double(dFdx(-1:2/10:1,y))
dFdX = 1x11
-6.9282 -5.5426 -4.1569 -2.7713 -1.3856 0 1.3856 2.7713 4.1569 5.5426 6.9282
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% Substitute numerical values in symbolic expression
U = subs(dFdx,[x y],{X,Y}); % call double to convert to numeric representation
V = subs(dFdy,[x y],{X,Y});
figure
fcontour(F,[-1 1],"Fill","on")
hold on
quiver(X,Y,U,V,'r')
hold off
And now compare with the numerical approach:
clear
% Numerical Solution
F= @(x,y) sqrt(3).*(2.*(x.^2+y.^2)-1);
fsurf(F,[-1 1])
[X, Y]=meshgrid(-1:2/10:1,-1:2/10:1);
figure
Z = F(X,Y);
surf(X,Y,Z)
% Numerical Gradient
[Fx,Fy] = gradient(Z);
figure
contourf(X,Y,Z)
hold on
quiver(X,Y,Fx,Fy,'r')
hold off

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更多回答（4 个）

Youssef Khmou
Youssef Khmou 2013-2-11
编辑：Youssef Khmou 2013-2-11
hi , you can use "gradient" :
[dF_x,dF_y]=gradient(F);
subplot(1,2,1), imagesc(dF_x), title(' dF(x,y)/dx')
subplot(1,2,2), imagesc(dF_y), title(' dF(x,y)/dy')
  2 个评论
Walter Roberson
Walter Roberson 2013-2-11
If you do not use the symbolic toolbox, gradient is numeric rather than analytic.
Youssef Khmou
Youssef Khmou 2013-2-11
编辑：Youssef Khmou 2013-2-11
True, but he has two sides because his example is numerical, you answered to the theoretical side ,while i answered to the numerical one,

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rapalli adarsh
rapalli adarsh 2019-1-9
syms c(x,y);
c(x,y)=input('enter cost Rs=\n');
cx=diff(c,x);
cy=diff(c,y);
s1=double(cx(80,20));
s2=double(cy(80,20));
if s1>s2 disp('fire standind stores')
else disp('fire standing stores')
end
Santhiya S
Santhiya S 2023-3-19
Using MATLAB, find the partial derivative with respect to ‘x’ and ‘y’ of the function f(x) = tan−1(x/y)
  1 个评论
Sergio E. Obando
Sergio E. Obando 2024-6-15
Ran in:
Replace your function in Walter's code:
syms f(x,y)
f(x,y) = atan(x/y)
f(x, y) = 
dFdx = diff(f,x)
dFdx(x, y) = 
dFdy = diff(f,y)
dFdy(x, y) = 

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Olivar Luis Eduardo
Good morning, I also have the same question, I have consulted a lot on the web, but they always give answers as if the surface were symbolic, but it is numerically and the calculation of the partial derivative of a matrix of order mxn remains.
  1 个评论
Sergio E. Obando
Sergio E. Obando 2024-6-15
编辑：Sergio E. Obando 2024-6-15
Please take a look at my comment above. The surface values are found by substituting/evaluating the symbolic expression at the grid points. Assuming you are using R2021b or later, you may find symmatrix useful for manipulation of matrix expressions, e.g. gradient of matrix multiplication

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