Using diff() or gradient() or other methods to solve the local slope of a set of discrete point?

Hi, I have a set of x and y, each has 44 values, and I use the following method to solve the local slope of this curve.

The two slopes look different, and I don't one which one is more reasonable?

load x.mat 
load y.mat
% 1. gradient, the result has 44 values (same with dataset)
slope1 = gradient(y); 
% 2. diff(), the result has 43 values
slope2 = diff(y)./diff(x);
% figure
plot(x,y,'ko-'); hold on;
plot(x,slope1,'r*-'); 
plot(x(2:end),slope2,'b*-'); 

Answer 1

[Jan]

load x.mat
load y.mat

slope1 = gradient(y); 
slope2 = diff(y) ./ diff(x);
slope3 = gradient(y, x);   % Consider x in the slope

plot(x, y, 'k-'); hold on;
plot(x, slope1, 'r-'); 
plot(x(2:end), slope2, 'b-'); 
plot(x, slope3, 'bo');

This shows, that diff(y)./diff(x) is a bestter choice except for the missing element, because diff(x) has one element less than x.

[Ying Wu]

Make more sense! Thanks!