How to calculate a numerical approximate derivative vector of a function?

I have a given formmula: Yprimenum(i) = (Y(i+1) – Y(i)) / ∆X, where ∆X is the X step length, or equivallently X(i) – X(i-1). And I also have two given functions: X= [0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 ] Y= [5 6 7 7.5 7.5 7.5 6.5 2.5 -5 -6 -6]

Now my task is to plot this function, Y, and calculate and plot the corresponding Yprimenum in the same graph. This is what I tried:

    x= [0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 ] 
y= [5 6 7 7.5 7.5 7.5 6.5 2.5 -5 -6 -6] 
yprime=2.*x; 
Yprimenum=zeros(1, length(x)-1); 
for i= 1:length(x)-1; 
Yprimenum(i)=(y(i+1)-y(i))./(x(i+1)-x(i)); 
end 
figure; 
hold on; 
plot(x,y); 
plot(x,yprime); 
plot(x,Yprimenum(i)); 
hold off; 
shg;

Answer 1

[Star Strider]

Your derivative, Yprimenum, is by definition one element shorter than x, so you have to eliminate the last entry of x to plot it:

x= [0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 ] 
y= [5 6 7 7.5 7.5 7.5 6.5 2.5 -5 -6 -6] 
yprime=2.*x; 
Yprimenum=zeros(1, length(x)-1); 
for i= 1:length(x)-1; 
    Yprimenum(i)=(y(i+1)-y(i))./(x(i+1)-x(i));
end 
figure; 
hold on; 
plot(x,y,'-b'); 
plot(x,yprime,'g'); 
plot(x(1:end-1),Yprimenum,'r'); 
hold off; 
shg;

This is unavoidable with the method you used (and that the diff function uses) but there are ways to deal with it. This is one such.

[Jabir Al Fatah]

perfect, thanks.

[Star Strider]

My pleasure!

Answer 2

[Andrei Bobrov]

x= [0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 ] 
y= [5 6 7 7.5 7.5 7.5 6.5 2.5 -5 -6 -6] 
Yprimenum = diff(y)./diff(x);

other variant

Yprimenum = gradient(y,x);