How can I draw the fitted curve and find the initial slope of the attached plot?

Hello, I am just beginner to matlab. Appriciate your help.

I attached the excel file for the x, y data.

Answer 1

[Star Strider]

Try this:

D = readmatrix('Example.xlsx');
x = D(:,1);
y = D(:,2);

p = polyfit(x, y, 3);
yfit = polyval(p, x);
dp = polyder(p);
dyfit = polyval(dp,x(1));
bint = (yfit(1) - dyfit*x(1));
xinitslope = [x(1)-x(75), x(1)+x(75)]; 
yinitslope = dyfit*xinitslope + bint;

figure
plot(x, y)
hold on
plot(x, yfit, 'LineWidth',1.5)
plot(xinitslope, yinitslope, 'LineWidth',3)
hold off
grid
legend('Data', '3\circ Polynomial Fit', 'Initial Slope', 'Location','N')
text(xinitslope(1), yinitslope(1), sprintf('\\leftarrow Initial Slope = %7.2f',dyfit), 'HorizontalAlignment','left', 'VerticalAlignment','middle')

producing:

.

[Selvarajah Premnath]

It's work ,Thanks a lot Star Strider.

Can you please advice how canwe choose the degree of polynomial,in my case 5th and 6th order alos fit those data and end up with diffrent slope values!

[Star Strider]

As always, my pleasure!

The choice of the degree of the polynomial is somewhat arbitrary. I chose 3 because I knew that 1 (linear) was inappropriate, and that 2 (quadratic) would be symmetrical, and the data are obviously not symmetrical. So 3 was an obvious simplest choice. A 7-degree polynomial is likely the highest reasonable order for any fit of any data.

So, use whatever gives the result you want, so long as the degree does not exceed 7.

A 3-degree fit gives the initial slope as -1320.15, and a 7-degree -1270.82. Calculating the percent difference between those as:

pct_dif = 100 * (-1320.15 - -1270.82)/mean([-1320.15 -1270.82])

yields:

pct_dif =
          3.80784030691209

so for all practical purposes, they are the same.

[Selvarajah Premnath]

Well explained.I got it. Thank you again.

[Star Strider]

As always, my pleasure!