How can i call an equation and it's derivative inside a matlab function?

As a newbie, i like to ask a simple question. I am trying to impliment a newton-rapson method for a simple equation as an example. I create a different matlab function from main function for the equation and call it inside the main function. However when I try to call the functions derivative it gives an error. I am aimin to not to take the derivative inside the main function for optimization concerns. I did try different methods but they give errors all the same.

function nr(x0,TC)

% TC is given in terms of percentage!

if nargin<2, x0=0; TC=10^-4;end

error=TC+1; i=0;

x(1)=x0;

while(error>TC)

x(i+2)=x(i+1)-f(x(i+1))/fd(x(i+1));

error=100*abs((x(i+2)-x(i+1))/x(i+2));

i=i+1;

end

fprintf('After %d iterations an approximate root is %f',i,x(i));

end

function [fx]=f(x)

fx=exp(-x)-x;

end

function fd=fd(x)

% syms x %These parts where i need help.

% fx=exp(-x)-x;

% fd=matlabFunction(diff(fx))

fd=-exp(-x)-1;

end

Answer 1

[Ömer Utku Örengül]

So I have it worked like this.

function fd=fd(a)
syms a
fx=exp(-a)-a;
fd=diff(fx);
end

and put "syms a" at the main function and finally changed "fd(x(i+1))" with;

...
x(i+2)=x(i+1)-f(x(i+1))/subs(fd,a,x(i+1));          %   subs(fd,a,x(i+1))
...

It worked as intended.

Answer 2

[Alan Stevens]

Like this, perhaps:

% TC is given in terms of percentage!
x0=0; TC=10^-4;
error=TC+1; i=0;
x(1)=x0;
while(error>TC)
    [fx, fd] = f(x(i+1));
    x(i+2)=x(i+1)-fx/fd;    
    error=100*abs((x(i+2)-x(i+1))/x(i+2));
    i=i+1;
end
fprintf('After %d iterations an approximate root is %f',i,x(i));

After 4 iterations an approximate root is 0.567143

function [fx, fd]=f(x)
fx=exp(-x)-x;
fd = -exp(-x)-1;
end

[Ömer Utku Örengül]

Yeah but i dont have problem with the main function. If i try to call the derivative like you did or i did (Giving the derivative equation by hand) there isn't any problem, howeverer if I try to cal it with matlab function diff it give errors.

function [fx]=f(x)
fx=exp(-x)-x;
end
function fd=fd(x)
% syms x                                                    %These parts where i need help.
% fx=exp(-x)-x;
% fd=matlabFunction(diff(fx))
fd=-exp(-x)-1;
end

Here if i try to use fd=diff(fx,x) matlab gives error. It works with the normal function but when I try to use the same method for it's derrivative it won't work. if i use syms it gaves dimension error etc.

[Walter Roberson]

There are two notable diff() functions. One of them only applies if the first parameter is symbolic or symbolic function.

syms x
fd = matlabFunction(diff(f(x),x))

fd = function_handle with value:
    @(x)-exp(-x)-1.0

function [fx]=f(x)
fx=exp(-x)-x;
end

[Alan Stevens]

You could always try something like this:

% TC is given in terms of percentage!
fx = @(x) exp(-x)-x;
dx = 10^-10;   % Choose a suitably small value
fd = @(x) (fx(x+dx) - fx(x))/dx;
x0=0; TC=10^-4;
error=TC+1; i=0;
x(1)=x0;
while(error>TC)
    x(i+2)=x(i+1)-fx(x(i+1))/fd(x(i+1));    
    error=100*abs((x(i+2)-x(i+1))/x(i+2));
    i=i+1;
end
fprintf('After %d iterations an approximate root is %f',i,x(i));

After 4 iterations an approximate root is 0.567143

but, if you have the Symbolic Maths package, Walter's suggestion is best.