Question about diff function

I have defined a function as follows

function A = A(V, h)
A = V/h + 2*sqrt(V*pi*h)
end

In another function, I try to call it:

h = h - A (V, h) / diff (A (V, h), h);

But I get the error

Error using / Matrix dimensions must agree.

I am at a loss. Can someone give me a hand?

[Matt J]

It is dangerous to use "A" as both a variable name and the name of your function.

Answer 1

[Star Strider]

To take a numerical derivative most accurately, use the gradient function.

[Francisco]

I was looking for an analytic derivative.

[Star Strider]

That requires the Symbolic Math Toolbox.

EDIT —

If you are satisfied with an approximation, you can do a numeric version of the definition of the derivative as an anonymous funciton:

dfdx = @(f,x)  (f(x+1E-8)-f(x))./1E-8;  % Conventional Derivative

A = @(V,h) V./h + 2*sqrt(V*pi*h);
V = 3;
h = linspace(0.1,10);
dAdh = dfdx(@(h) A(V,h), h);

figure(1)
plot(h, A(V,h), '-b')
hold on
plot(h, dAdh, '-r')
hold off
legend('A(V,h)', 'dA(V,h)/dh')

You may want to experiment with the 1E-8 value (this is the value that most often ‘works’ for me). At the very least, it will provide you with something to experiment with.

Answer 2

[Roger Stafford]

   h = h - (V/h+2*sqrt(V*pi*h))/(-V/h^2+V*pi/sqrt(V*pi*h));

(Maybe I am old-fashioned, but with elementary functions like this, it's easier to use calculus than to have to mess around with the 'diff' function in my opinion.)

Answer 3

[Azzi Abdelmalek]

a=[ 2  4  8]
b=diff(a)

b= [4-2  8-4]

a is 1x3 and b is 1x2

[Francisco]

I thought diff was http://www.mathworks.com/help/symbolic/diff.html, as in, I was trying to differentiate A. How can I achieve this, then?

[Azzi Abdelmalek]

B=A(V,h)
C=diff (A (V, h), h);
h = h - B(h+1:end)/C ;