How to I compute partial derivatives of a function

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Mat Hunt
Mat Hunt 2016-8-11
评论： John D'Errico 2016-8-12
Suppose I have a function z=z(x,y), how do I numerically (not symbolically) compute the partial derivatives?
I know of the function gradient(f,dx) which computes general derivatives in one dimension, but what is I want to compute the function:
\frac{\partial^{4}z}{\partial x^{4}}+\frac{\partial^{2}z}{\partial y^{2}}
for example? So I would need to compute them separately. I would rather not do a finite difference solution as that would be a faff. Is there a way of using the gradient function at all?

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回答（2 个）

Torsten
Torsten 2016-8-12
Compute the derivatives symbolically using "diff" and turn the result in a function handle using "matlabFunction".
Best wishes
Torsten.
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Mat Hunt
Mat Hunt 2016-8-12
No, mixed derivatives are not required this time, but I need to calculate a sixth order derivative in x and a second order derivative in y.
If I arrange Z as a meshgrid, I can look at doing gradient on separate rows and columns I suppose.
Torsten
Torsten 2016-8-12
Yes, exactly, you will have to loop over the rows or columns of the z-matrix.
Best wishes
Torsten.

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John D'Errico
John D'Errico 2016-8-12
High order partials can be difficult to estimate numerically, and to do so with full precision.
The tool derivest (found on the file exchange) can do a decent job though.
http://www.mathworks.com/matlabcentral/fileexchange/13490-adaptive-robust-numerical-differentiation
Consider this example function:
z = @(x,y) exp(-(x+2*y).^2);
z(-1,1)
ans =
0.36788
I'll define the variables x0 and y0 so that you can see how to use it. So we want to compute the 4 order partials around the point (x0,y0).
x0 = -1;
y0 = 1;
[dzdx4,ex] = derivest(@(x) z(x,y0),x0,'deriv',4)
dzdx4 =
-7.3576
ex =
3.4866e-08
[dzdy4,ey] = derivest(@(y) z(x0,y),y0,'deriv',4)
dzdy4 =
-117.72
ey =
1.7325e-06
The second returned argument is an error estimate that indicates how well it thinks it did the job. So I am getting roughly 8 significant digits of precision in each direction.
I did them separately before to see the error estimates also.
In one line do this:
D = derivest(@(x) z(x,y0),x0,'deriv',4) + ...
derivest(@(y) z(x0,y),y0,'deriv',4)
D =
-125.08
  2 个评论
Mat Hunt
Mat Hunt 2016-8-12
Hi John,
I am solving a PDE using the Newton method, so my function isn't symbolic, it's just a series of numbers (for ease I am considering writing the matrix as a vector), so I can't write it as a function as it's technically a variable. So I don't know if I can write it as function handle.
John D'Errico
John D'Errico 2016-8-12
I NEVER said the problem needed to be symbolic, did I? In the example I showed, nothing was symbolic, just a function, z(x,y), as you said that you had. But you never said that all you really have is a series of numbers. Should I have known that? Oh, well. No. You cannot use derivest.
If you have no more than a list of numbers, then you need to generally need to use a finite difference approximation. Or you can use finite elements. There are lots of classic ways to solve PDES. Books of them, even.

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