How to I compute partial derivatives of a function

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?

Answer 1

[Torsten]

Compute the derivatives symbolically using "diff" and turn the result in a function handle using "matlabFunction".

Best wishes

Torsten.

[Mat Hunt]

Hi, you didn't see the thing about not symbolically did you?

Let be make it more clear for you. Suppose I have the code:

x=linspace(-6,6,30);y=linspace(-6,6,30);
[X,Y]=meshgrid(x,y);

Then if I have a function defined numerically, how do I compute dz_dx and dz_dy separately and higher derivatives.

[Torsten]

You wrote you have a function z=z(x,y) - so I assumed the function is given as an analytical formula.

If you only have values of the function on a stuctured grid, there is no other way than to use finite difference approximations for the partial derivatives, I guess.

Best wishes

Torsten.

[Mat Hunt]

In 1D I simply used gradient(f,dx), is there no higher genaralisation to 2D or 3D?

Mat

[Torsten]

Do you want to calculate mixed derivatives, e.g. d^2z/dxdy ?

As long as this is not the case, the "gradient" function should suffice also to compute higher-order derivatives. E.g.

d^2z/dx^2 = gradient(gradient(z,dx),dx)

Best wishes

Torsten.

[Mat Hunt]

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]

Yes, exactly, you will have to loop over the rows or columns of the z-matrix.

Best wishes

Torsten.

Answer 2

[John D'Errico]

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

[Mat Hunt]

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]

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.