how to calculate the partial derivatives for a given function of two variable ?

How can I write code to calculate the partial derivatives a given function of two variable ?

Answer 1

[Dyuman Joshi]

Hello, You can use diff function operator to obtain partial derivatives as follows:

1- Define the function using symbolic variables

%Random function for example
syms x y;
f(x,y) =  x^2 + y^2 + x*y;

2-use diff with respect to the variable you want to differentiate.

fx = diff(f,x)

fx(x, y) = 

fy = diff(f,y)

fy(x, y) = 

You can also find the value of parial derivatives by substituting the value -

fx(1,0)

ans = 

2

Answer 2

[IBTASHAM IMAM]

Using MATLAB, find the partial derivatives of F(x,y)=x^3+y^3+6xy-1 with respect to y at the point (1,1).

[IBTASHAM IMAM]

% Symbolic variables
syms x y lambda;
% Define the function and constraint
f = 3*x + 4*y;
g = x^2 + y^2 - 1;
% Define the Lagrange function
L = f + lambda*g;
% Take partial derivatives of L with respect to x, y, and lambda
Lx = diff(L, x);
Ly = diff(L, y);
Ll = diff(L, lambda);
% Solve the system of equations for critical points
eqn1 = Lx == 0;
eqn2 = Ly == 0;
eqn3 = Ll == 0;
solutions = solve([eqn1, eqn2, eqn3], [x, y, lambda]);
% Extract the solutions
x_sol = solutions.x;
y_sol = solutions.y;
lambda_sol = solutions.lambda;
% Substitute critical points back into the original constraint
g_val = subs(g, {x, y}, {x_sol, y_sol});
% Check for valid solutions (points on the circle)
valid_solutions = abs(g_val) < 1e-10; % Allow for numerical precision issues
% Extract valid critical points and function values
x_valid = double(x_sol(valid_solutions));
y_valid = double(y_sol(valid_solutions));
f_val = double(subs(f, {x, y}, {x_valid, y_valid}));
% Find the maximum and minimum values of f
f_max = max(f_val);
f_min = min(f_val);
% Display results
disp('Maximum value of f(x, y):');
disp(f_max);
disp('Minimum value of f(x, y):');
disp(f_min);