# partial derivatives matrix and Integrations

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Eddy Ramirez on 6 Oct 2021
Answered: Abhishek Kolla on 11 Nov 2021
Kindly requesting your assistance, I am running the following code and I am getting some errors. I am trying to get the partial derivative of a matrix in terms of X and Y. Based on the research I have conducted, it says that I could utilize the gradient option, but the error says "too many output arguments".
Also, for integrations would it be possible to get the feedback if I am coding it correctly. The X integration is from 0 to 4 and y from 0 to 3. I ran the code, but since I am getting the error for the partial derivative the code will not run the last part.
E=30*10^6;
v=.3;
t=.1; %inches - Thickness
a=4/2;
b=3/2;
x1=0;
x2=4;
x3=4;
x4=0;
y1=0;
y2=0;
y3=3;
y4=3;
syms x
syms y
D=(E/(1-v^2))*[1 v 0;v 1 0; 0 0 ((1-v)/2)];
h1=(1/(4*a*b))*(x-x2)*(y-y4);
h2=-(1/(4*a*b))*(x-x1)*(y-y3);
h3=(1/(4*a*b))*(x-x4)*(y-y2);
h4=-(1/(4*a*b))*(x-x3)*(y-y1);
H=[h1 0 h2 0 h3 0 h4 0; %X
0 h1 0 h2 0 h3 0 h4]; %Y
B=D*[HX,HY]*t;
K=(int(int(B,x,0,4),y,0,3))

Abhishek Kolla on 11 Nov 2021
Hi,
In the above code the output of gradient will return x and y components of the two dimensional numerical gradient of matrix F. More detailed explanation on gradient can be found here Numerical gradient - MATLAB gradient (mathworks.com)
Define any interval range for both x and y after declaring x and y
x = -2:0.2:2;
y = x';
After making the following changes the gradient function will work and both HX,HY will be obtained