How to evaluate symbolic derivative?

I have created a function, powerdiff, but when I try to evaluate it at a number I get the following error message. I am wanting to evaluate it at different points to try and find the root:

Subscript indices must either be real positive integers or logicals.

Error in sym/subsref (line 805)
            R_tilde = builtin('subsref',L_tilde,Idx);

Error in ivcurve_var_zoom (line 36)
powerdiff(1e-2)

Here is my code. I can evaluate the original function, power, just not powerdiff (the derivative).

syms J
Veff=@(J)0;
%Creating voltage equation 
for i=1:numbg
   Jph=flux(i,2);%Constant
   Avalue=A(i);%Constant

     V=@(J)k*t*log((Jph-J)/Avalue)+bg(i);%Voltage function
     Veff=@(J)Veff(J)+V(J);%Adds to previous voltage function
  end

power=@(J)Veff(J).*J;
power(1e-2)
powerdiff=diff(power(J));
powerdiff(1e-2)

Answer 1

[Stephan]

Hi,

you are treating the derivate like the function handle - but it is not a function handle.

What you do is asking matlab for the value of the vector powerdiff at the index 1e-2...

What you probably want to do is get the result of that function for the input value 1e-2.

You could use matlabFunction to achieve what you want.

Best regards

Stephan

[Ronan Garrison]

Thank you Stepahn, matlabFunction works perfectly! Now I can use fzero but it isn't working seeing as my function has an asymptote (numbers become imaginary). How can i determine where the asymptote of a function is so I can put in the appropriate limit in fzero?

[Stephan]

https://de.mathworks.com/help/symbolic/find-asymptotes-critical-and-inflection-points.html

[Ronan Garrison]

My asymptote is a vertical asymptote... This would only work for a horizontal asymptote... Here is an image of my plot. I need to determine what x value the dashed line falls at, so I can use this as the limit for fzero.

[Stephan]

Is not there a section about finding vertical asymptots directly below the section about horizontal asymptots?

[Ronan Garrison]

Yes but roots does not work well with symbolic expressions...

[Stephan]

Can you provide the symbolic function of powerdiff ?

[Ronan Garrison]

@(J)log(J.*(-3.937007874015748e-4)+2.442872757695621e-6).*2.58519909e-2+log(J.*(-6.835083114610674e-3)+3.826599732420725e-4).*2.58519909e-2+J.*(1.017794917322835e-5./(J.*3.937007874015748e-4-2.442872757695621e-6)+1.767005064796588e-4./(J.*6.835083114610674e-3-3.826599732420725e-4))+3.1e1./1.0e1

It is a big function that will change, but keep the same general structure.

[Stephan]

Are these values possible candidats:

x01 =

     0.006204896804547

x02 =

     0.055984684725208

???

What did i do?:

% set J to symbolic variable
syms J
% Your function
f = log(J.*(-3.937007874015748e-4)+2.442872757695621e-6).*2.58519909e-2+log(J.*(-6.835083114610674e-3)+3.826599732420725e-4).*2.58519909e-2+J.*(1.017794917322835e-5./(J.*3.937007874015748e-4-2.442872757695621e-6)+1.767005064796588e-4./(J.*6.835083114610674e-3-3.826599732420725e-4))+3.1e1./1.0e1;
% divide in numerator and denominator 
[num, den] = numden(f);
% set denominator equal to zero
f_asymp_vert = den == 0;
% solve this equation
x0 = solve(f_asymp_vert);
% set format
format long;
% extract the (still) symbolic values of the solution and convert to double
x01 = double(x0(1,1))  % --> 1831362293716070975/295147905179352825856
x02 = double(x0(2,1))  % --> 252132605266902219/4503599627370496000

--> Is that what you need?

Best regards

Stephan