Newtons method for bivariate function Ive managed to comple

Newton\'s method for bivariate function - I\'ve managed to complexify the equation, and break down f to f1 & f2 where f1(x,y)=x^2-y^2+1 & f2(x,y)=2xyi. Hoping for insight on how to run newton\'s method and plot the function

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#include<stdio.h>

#include<math.h>

float f(float x)

{

return x*log10(x) - 1.2;

}

float df (float x)

{

return log10(x) + 0.43429;

}

void main()

{

int itr, maxmitr;

float h, x0, x1, allerr;

printf(\"\ Enter x0, allowed error and maximum iterations\ \");

scanf(\"%f %f %d\", &x0, &allerr, &maxmitr);

for (itr=1; itr<=maxmitr; itr++)

{

h=f(x0)/df(x0);

x1=x0-h;

printf(\" At Iteration no. %3d, x = %9.6f\ \", itr, x1);

if (fabs(h) < allerr)

{

printf(\"After %3d iterations, root = %8.6f\ \", itr, x1);

return 0;

}

x0=x1;

}

printf(\" The required solution does not converge or iterations are insufficient\ \");

return 1;

}

Newton\'s method for bivariate function - I\'ve managed to complexify the equation, and break down f to f1 & f2 where f1(x,y)=x^2-y^2+1 & f2(x,y)=2xyi.
Newton\'s method for bivariate function - I\'ve managed to complexify the equation, and break down f to f1 & f2 where f1(x,y)=x^2-y^2+1 & f2(x,y)=2xyi.

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