To take the partial derivative of a function using matlab

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!

Answer 1

[Walter Roberson]

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)

[Pranjal Pathak]

Thank you sir for your answers. Actually I need the analytical derivative of the function and the value of it at each point in the defined range. i.e. diff (F,X)=4*3^(1/2)*X; is giving me the analytical derivative of the function. After finding this I also need to find its value at each point of X( i.e., for X=(-1:2/511:+1). Similarly the others.

[Walter Roberson]

dX = matlabFunction(diff(F,X));
dY = matlabFunction(diff(F,Y));
[x, y]=meshgrid(-1:2/511:+1, -1:2/511:+1);
dX(x,y)
dY(x,y)

[Palak Kapoor]

Error: Too many input arguments.

[Olivar Luis Eduardo]

I think that numerical calculation is being requested, not the symbolic one. In other words, the surface is a matrix

[Sergio E. Obando]

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

Answer 2

[Youssef Khmou]

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')

[Walter Roberson]

If you do not use the symbolic toolbox, gradient is numeric rather than analytic.

[Youssef Khmou]

True, but he has two sides because his example is numerical, you answered to the theoretical side ,while i answered to the numerical one,

Answer 3

[rapalli adarsh]

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

Answer 4

[Santhiya S]

Using MATLAB, find the partial derivative with respect to ‘x’ and ‘y’ of the function f(x) = tan−1(x/y)

[Sergio E. Obando]

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) = 

Answer 5

[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.

[Sergio E. Obando]

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