# Problem in evaluating the value of the derivative of multivariable function

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Galar on 24 Sep 2022
Commented: Galar on 24 Sep 2022
I'm triyng to make a function that uses the Newton - Raphson method to solve a system of 2 equation
function [x_final, y_final, f1_final, f2_final, iter, errs, arrs] = Sys_equa_newton(f1, f2, x0, y0, err, max_iter)
%newton_sys - Description
%
% Syntax: y = newton_sys(input)
%
% Long description
% f1: first funtion of 2 variables
% f2: second funtion of 2 variables
% x0: initial value of x
% y0: initial value of y
% err: threshold value of error to stop
% max_iter: maximum iteration to stop
cur_err = 1;
iter = 0;
syms x y;
df1_x = matlabFunction(diff(f1, x));
df1_y = matlabFunction(diff(f1, y));
df2_x = matlabFunction(diff(f2, x));
df2_y = matlabFunction(diff(f2, y));
x = x0;
y = y0;
X_n = f1(x, y) * (df2_y(x, y)) - f2(x, y) * (df1_y(x, y));
Y_n = f2(x, y) * (df1_x(x, y)) - f1(x, y) * (df2_x(x, y));
Deno = df1_x(x, y) * df2_y(x, y) - df1_y(x, y) * df2_x(x, y);
arrs = zeros(2,max_iter);
errs = zeros(2,max_iter);
while ((cur_err > err) && (iter <= max_iter))
x_old = x;
x = x_old - X_n / Deno;
y_old = y;
y = y_old - Y_n / Deno;
err_x = abs(x - x_old) / x;
err_y = abs(y - y_old) / y;
arrs(:,iter+1) = [x; y];
errs(:,iter+1) = [err_x; err_y];
iter = iter+1;
cur_err = err_x + err_y;
end
x_final = x;
y_final = y;
f1_final = f1(x_final,y_final);
f2_final = f2(x_final,y_final);
end
But if I enter a function that only has the first power of x or y or both x and y are in first power, when i calculate the derivatrive (df1_x for example) then the new function only have 1 argument or 0 argument if both x and y are in first power.
So is there anyway to solve this problem, I'm trying to make a general function that can work with all 2 system of equation

Walter Roberson on 24 Sep 2022
matlabFunction(diff(f1, x), 'vars', [x, y] )
Galar on 24 Sep 2022
Thank you, this works perfectly for me

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