Differentiation of two variables in a polynomial equation

Please the differential part gives giving me empty curly brackets instead of values in this code set.

% Initial x1 and x2
x1 = 5, x2 = 4
% Equations
y1 = (x1-3)^2 + (2*x2 - 8)^2
y2 =(x1-4)^2 + (x2 - 10)^2
%Given y values
y = [5;49]
% First set of y values
yo = [y1;y2]
f = y - yo
A11 = diff(poly((x1-3)^2 + (2*x2 - 8)^2),x1)
A12 =diff(poly((x1-3)^2 + (2*x2 - 8)^2),x2)
A21 =diff(poly((x1-4)^2 + (x2 - 10)^2),x1)
A22 =diff(poly((x1-4)^2 + (x2 - 10)^2),x2)

| I was expecting to see this: | http://www.mathworks.com/matlabcentral/answers/uploaded_files/25540/Capture.JPG

or http://www.mathworks.com/matlabcentral/answers/uploaded_files/25542/Capture1.JPG

Thank you.

Answer 1

[John D'Errico]

x1 and x2 are numbers. You set them as such. Remember this line?

x1 = 5, x2 = 4

you cannot differentiate with respect to a number.

[bayoishola20]

okay, then what should it be? Because when I do this:

A11 = diff(poly((x1-3)^2 + (2*x2 - 8)^2),x1)

It tells me

Undefined function or variable 'x1'.

[John D'Errico]

I'd start with the statement

syms x1 x2

This tells MATLAB that these are symbolic variables, something that you can differentiate.

When you defined them instead as 5 and 4 respectively, you told MATLAB they were simple numbers, just constants.

Later on, IF you have a specific value to evaluate a derivative at, then you can use subs.

Oh, by the way, When you define the variable A11, this does not create the (1,1) element of the matrix A, just a scalar variable named A11. I think you need to learn about brackets, [], and how to concatenate things into an array. Thus, see what this does:

[1 2;3 4]