x values when taking a numerical derivative

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I want to calculate the first derivative f(x) = dy/dx for data that is irregularly spaced in x. I think (correct me if I'm wrong) this can be done by diff(y)./diff(x)
But the resulting vector is one less than the length of the x and y vectors, so what are the corresponding x values? I want to then calculate x dy/dx, so it would be helpful to know what to use.

Risposta accettata

Star Strider
Star Strider il 6 Feb 2023
I think (correct me if I'm wrong) this can be done by diff(y)./diff(x)
Not actually wrong, simply mistaken with respect to expecting that the result of diff will be what you want.
Use the gradient function (or Torsten’s approach, that does essentially the same operation) instead:
dydx = gradient(y) ./ gradient(x);
.
  6 Commenti
Torsten
Torsten il 6 Feb 2023
the support page for gradient says that FX = gradient(F) corresponds to /, so why do you recommend dividing by the gradient in x in this case?
Because in the case FX = gradient(F), the distance between the x values is assumed to be 1 for all of them.
Or do you think MATLAB knows how your x vector looks like if you don't supply it ?

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Più risposte (4)

Sulaymon Eshkabilov
Sulaymon Eshkabilov il 6 Feb 2023
Yes, you are right, e.g.:
x = [0 .2 .3 .45 .65 .75 .96]
x = 1×7
0 0.2000 0.3000 0.4500 0.6500 0.7500 0.9600
y = [-3 10 11 12 13 12 9]
y = 1×7
-3 10 11 12 13 12 9
dydx = diff(y)./diff(x)
dydx = 1×6
65.0000 10.0000 6.6667 5.0000 -10.0000 -14.2857
yyaxis left
plot(x, y, 'b-o', 'MarkerFaceColor', 'y', 'DisplayName', 'y(x)')
ylabel('y(x)')
yyaxis right
plot(x(1:end-1), dydx, 'r--p', 'MarkerFaceColor','c','DisplayName', 'dy/dx')
ylabel('dy/dx')
xlabel('x')
grid on
legend show
  2 Commenti
L'O.G.
L'O.G. il 6 Feb 2023
Modificato: L'O.G. il 6 Feb 2023
Thanks. So just to be clear, why does dydx(end-1) correspond to x(end)?
Sulaymon Eshkabilov
Sulaymon Eshkabilov il 6 Feb 2023
difference1 is between 1 and 2 and then 2 and 3, etc., and thus end-1

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Tushar Behera
Tushar Behera il 6 Feb 2023
Modificato: Tushar Behera il 6 Feb 2023
Hi L'O.G,
I believe you want to calculate derivate for two separate datasets. In order to do that you can use 'diff(y)./diff(x)' for example:
clc;
clear;
close all;
x=linspace(0,2*pi,100);
y=sin(x);
yp=cos(x);
dx=diff(x);
dy=diff(y);
yp_hat=dy./dx;
err=yp(1:end-1)-yp_hat;
figure;
subplot(1,2,1);
plot(x,y);
hold on;
plot(x(1:end-1),yp_hat)
xlabel('x');
ylabel('y');
legend('original function','Approx derivative');
grid on;
subplot(1,2,2);
plot(x(1:end-1),err);
xlabel('x');
ylabel('error');
here 'x' and 'y' are two vectors and by using 'diff(y)./diff(x)' you can calculate the first order derivate which is 'cos(x)'. To answer the question which values of x corresponds to which 'yp_hat' . you can get that by using,
x(1:end-1)
I hope this solves your query.
Regards,
Tushar

Torsten
Torsten il 6 Feb 2023
Modificato: Torsten il 6 Feb 2023
Use
n = numel(y);
dydx(1) = (y(2) - y(1))/(x(2) - x(1));
dydx(2:n-1) = (y(3:n) - y(1:n-2))./(x(3:n) - x(1:n-2));
dydx(n) = (y(n) - y(n-1))/(x(n) - x(n-1));

Sulaymon Eshkabilov
Sulaymon Eshkabilov il 6 Feb 2023
Modificato: Sulaymon Eshkabilov il 6 Feb 2023
There will be some significantly different results from diff() and gradient() if the increment of x varies. See this simulation:
x = [0 .2 .3 .45 .65 .75 .96];
y = [-3 10 11 12 13 12 9];
dy1 = diff(y)./diff(x);
dy2 = gradient(y)./gradient(x);
for ii = 1:length(x)-1
dy3(ii) = (y(ii+1)-y(ii))/(x(ii+1)-x(ii));
end
N = numel(y);
dy4(1) = (y(2)-y(1))/(x(2)-x(1));
dy4(2:N-1) = (y(3:N)-y(2:N-1))./(x(3:end)-x(2:end-1));
dy4(N) = (y(end)-y(end-1))/(x(end)-x(end-1));
plot(x(1:end-1), dy1, 'b-o', 'MarkerFaceColor', 'y', 'DisplayName', 'diff', 'markersize', 13)
hold on
plot(x(1:end), dy2, 'rs--', 'MarkerFaceColor', 'c', 'DisplayName', 'gradient')
plot(x(1:end-1), dy3, 'g--p', 'MarkerFaceColor','y','DisplayName', 'Loop computed difference', 'MarkerSize', 10)
hold on
plot(x(1:end), dy4, 'k--h', 'MarkerFaceColor','c','DisplayName', 'vectorized: gradient')
ylabel('dy/dx')
xlabel('x')
grid on
legend show
% Note that as the increment of x gets smaller the error (offset) will also
% diminish. See this example: dx = 0.063467 vs. dx = 0.006289:
x=linspace(0,2*pi,100);
dx = x(2);
y=sin(x);
ANS=cos(x);
dY1=diff(y)./diff(x);
dY2 = gradient(y)./gradient(x);
for ii = 1:length(x)-1
dY3(ii) = (y(ii+1)-y(ii))/(x(ii+1)-x(ii)); % The same as diff()
end
n = numel(y);
dY4(1) = (y(2)-y(1))/(x(2)-x(1));
dY4(2:n-1) = (y(3:end)-y(1:end-2))./(x(3:end)-x(1:end-2));
dY4(n) = (y(end)-y(end-1))/(x(end)-x(end-1));
E1=ANS(1:end-1)-dY1;
E2 = ANS-dY2;
E3 = ANS(1:end-1)-dY3;
E4 = ANS-dY4;
fprintf(['Norm of errors @ dx = %f: ' ...
'E_diff = %f; E_gradient = %f; ' ...
'E_loop = %f; E_grad_vect = %f \n'], [dx, norm(E1) norm(E2) norm(E3) norm(E4)])
Norm of errors @ dx = 0.063467: E_diff = 0.223238; E_gradient = 0.004770; E_loop = 0.223238; E_grad_vect = 0.004770
% 10 times smaller incremental step of x than the previous example leads to
% the reduction of error norm to more than 3 times
x=linspace(0,2*pi,1000);
dx = x(2);
y=sin(x);
ANS=cos(x);
dY1=diff(y)./diff(x);
dY2 = gradient(y)./gradient(x);
for ii = 1:length(x)-1
dY3(ii) = (y(ii+1)-y(ii))/(x(ii+1)-x(ii)); % The same as diff()
end
n = numel(y);
dY4(1) = (y(2)-y(1))/(x(2)-x(1));
dY4(2:n-1) = (y(3:end)-y(1:end-2))./(x(3:end)-x(1:end-2));
dY4(n) = (y(end)-y(end-1))/(x(end)-x(end-1));
E1=ANS(1:end-1)-dY1;
E2 = ANS-dY2;
E3 = ANS(1:end-1)-dY3;
E4 = ANS-dY4;
fprintf(['Norm of errors @ dx = %f: ' ...
'E_diff = %f; E_gradient = %f; ' ...
'E_loop = %f; E_grad_vect = %f \n'], [dx, norm(E1) norm(E2) norm(E3) norm(E4)])
Norm of errors @ dx = 0.006289: E_diff = 0.070283; E_gradient = 0.000147; E_loop = 0.070283; E_grad_vect = 0.000147

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