Strange slope for points in a curve

Dear experts, I want to get the slope for each point from the curve defined below:

x = 1194:1206;
y =[114,115,117,118,119,120,122,123,124,126,127,128,129];

see the attached graph for the curve.

Directly applying the function "gradient" cannt get the correct slope - the slope will jump up an down. It's understandable that I don't have enough data points. I expect the slope changes gradually.

I expect not to use any interpretation or curve fitting method. Any suggestion is welcome. Many thanks

Answer 1

[Mohammad Abouali]

There are two slopes at each point. The left slope and right slope? What you want to do about that?

[Kyle]

could you please suggest to calculate the right slope?

[Mohammad Abouali]

It gonna still jumps up and down since your lines are not straight any way.

If you want first order accurate first derivatives do this

x = 1194:1206;
y =[114,115,117,118,119,120,122,123,124,126,127,128,129];
dYO1=(y(2:end)-y(1:end-1))./(x(2:end)-x(1:end-1)); % Or in this case it would be the same as diff(y);
figure, plot(x(1:end-1),dYO1,'o-');

You can calculate second order accurate first derivative as follow:

dYO2=(y(3:end)-y(1:end-2))./(x(3:end)-x(1:end-2));
figure, plot(x(2:end-1),dYO2,'ro-')

if you want the slope of the lines themselves (i.e. at the middle of the line) then dYO1 would be the accurate one ( at the center of the lines not at the point themselves)

xc=0.5*(x(1:end-1)+x(2:end));
plot(xc,dYo1);

Answer 2

[Matt J]

You cannot define a derivative without some sort of interpolation model. You can take first differences using diff(y), but that corresponds to a linear interpolation model.

[Kyle]

I tried using spline() function. Although it enhances the smoothness, the slope vs x still jumps up and down :(

[Matt J]

The slope is obviously going to jump up and down no matter what you do. Your points don't all lie on a straight line.