If I had the derivative of an equation and had data points how would I plot them

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Batuhan Yildiz
Batuhan Yildiz 2022-10-2
回答： Walter Roberson 2022-10-2
My derivative is 1 - (sin(x))/(2√(x)) - (√x)(cos(x)) and x = [1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0] how would I plot them?

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采纳的回答

Torsten
Torsten 2022-10-2
Ran in:
d = @(x)1 - sin(x)./(2*sqrt(x)) - sqrt(x).*cos(x);
x = [1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0];
plot(x,d(x))
  2 个评论
Batuhan Yildiz
Batuhan Yildiz 2022-10-2
编辑：Image Analyst 2022-10-2
Ran in:
Two Questions:
  1. why add the @(x)?
  2. would this code also work?
x=[1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0] ;
y=x-sqrt(x).*sin(x);
dy=diff(y)./diff(x);
plot(x(2:end),dy);
Torsten
Torsten 2022-10-2
编辑：Torsten 2022-10-2
Ran in:
why add the @(x)?
I defined the derivative d as a function. Alternatively, you can of course use
x = [1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0];
d = 1 - sin(x)./(2*sqrt(x)) - sqrt(x).*cos(x);
plot(x,d)
would this code also work?
If you are too lazy to compute the exact derivative, you can use the approximation
dy = diff(y)./diff(x)
formed by finite difference quotients.
But note that d gives you an approximate derivative in points (x(1:end-1)+x(2:end))/2, not in x(2:end). Thus for the plot you should use
plot((x(1:end-1)+x(2:end))/2,dy)
instead of
plot(x(2:end),dy)
Alternatively, you can compute the derivative symbolically:
syms x
y = x-sqrt(x).*sin(x);
dy = diff(y,x)
dy = 

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更多回答（1 个）

Walter Roberson
Walter Roberson 2022-10-2
if you have numeric derivative and you want to plot the original curve (to within a constant shift) then use cumtrapz() and plot that.

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