Compute derivative of spline function at a certian value

If I have x,y data, and I spline interpolate them:

sp=spline(x,y);
D=@(x) ppval(sp,x);

I want to compute the value of the derivative of the spline function at specific x values, say x1,x2.

[darova]

What about numerical approach? diff maybe?

Answer 1

[John D'Errico]

By far the simplest is to use fnder, IF you have the curve fitting toolbox.

x = linspace(-pi,pi,100);
y = sin(x);
f = spline(x,y);
dfdx = fnder(f);
fnplt(dfdx)

Of course the derivative function should be -cos(x), which is clearly well-approximated here.

You can now evaluate the derivative function at a point easily enough using either ppval or fnval.

ppval(dfdx,pi/4)
ans =
      0.70711
      
fnval(dfdx,pi/4)
ans =
      0.70711

If you lack the curve fitting toolbox (get it, if you will be doing any curve fitting at all) you can still do the differentiation easily enough.

D = [3 0 0 0;0 2 0 0;0 0 1 0]';
fp = f;
fp.order = 3;
fp.coefs = fp.coefs*D;

Now to test it...

ppval(fp,pi/4)
ans =
      0.70711

So the differentiation was quite easy, even without fnder.

[Amdad Chowdury]

Thats really helps. Thank you.

Answer 2

[Ameer Hamza]

Something like this

x = 1:10;
y = x.^2;
sp = spline(x, y);

x1 = 2.5; % point to calculate the derivative
reg = find(x1 < sp.breaks, 1)-1;
deri_val = polyval(polyder(sp.coefs(reg, :)), x1-sp.breaks(reg))

Or create a function handle

x = 1:10;
y = x.^2;
sp = spline(x, y);

reg = @(x1) find(x1 < sp.breaks, 1)-1;
deri_val = @(x1) polyval(polyder(sp.coefs(reg(x1), :)), x1-sp.breaks(reg(x1)));

Result

>> deri_val(2)
ans =
    4.0000
>> deri_val(2.5)
ans =
     5
>> deri_val(7)
ans =
   14.0000