My diff function won't work and I'm not sure why

My goal is to give an initial x, and return the function values at that x

function [f, g, h] = myfunction(x)
    syms x
    q = exp(2*sin(x)) - x;
    f = q(x);
    dq = diff(q);
    g = dq(x);
    dq2 = diff(dq);
    h = dq2(x);
end

So what I've been struggling with is that if I assign my equation as above, then I can find "f = q(x)" but there is an error for the "diff". If I change the syntax to

q = @(x) exp(2*sin(x)) - x;

then I'm able to calculate using "diff" but then the "f = q(x)" doesn't work. I'm pretty new to MATLAB and not great at it so any advice would be great!

Update: This is the error message I am recieving for the first code

Error using sym/subsindex (line 855)

Invalid indexing or function definition. Indexing must follow MATLAB indexing. Function arguments must be symbolic variables, and function body must be sym expression.

Error in sym/subsref (line 900)

R_tilde = builtin('subsref',L_tilde,Idx);

Error in myfunction (line 4)

f = q(x);

And for the other code it is:

Undefined function 'diff' for input arguments of type 'function_handle'.

Error in myfunction2 (line 4)

dq = diff(q);

[Steven Lord]

What is the full and exact text of the error messages you receive in each of the circumstances? Please show us all the text displayed in red and/or orange in the Command Window when you try to run this code.

Answer 1

[Dimitris Kalogiros]

Try this :

clearvars; close all; clc;

syms x
q(x) = exp(2*sin(x)) - x;
f(x) = q(x)
dq(x) = diff(q(x));
g(x) = dq(x)
dq2(x) = diff(dq(x));
h(x) = dq2(x)

or this one:

clearvars; close all; clc;

syms x
q(x) = exp(2*sin(x)) - x;
f = q(x)
dq(x) = diff(q(x));
g = dq(x)
dq2(x) = diff(dq(x));
h = dq2(x)

[Sarah Johnson]

Both of these still are only producing the functions themselves and not the value. I'm not sure what's wrong.

Answer 2

[David Goodmanson]

Hi Julia,

if you want to use this for a variety of functions, then

fun = @(x) exp(2*sin(x)) - x;        % one example

[f g h] = newfuns(fun)

function [a b c] = newfuns(fun)
    syms x
    a = fun(x);
    b = diff(a);
    c = diff(b);
end

f =
 exp(2*sin(x)) - x
 
g =
 2*exp(2*sin(x))*cos(x) - 1
 
h = 
4*exp(2*sin(x))*cos(x)^2 - 2*exp(2*sin(x))*sin(x)

[Sarah Johnson]

My goal is to find the function values at the initally given x. Not find the differentials themselves. I have done that but I'm having trouble plugging in the inital x to give me an answer. For example, if I plug in x = 0 and then call "myfunction(x)", I would want my f to be 1

Answer 3

[Dimitris Kalogiros]

I think , the answer to your problem is the following piece of code:

clearvars; clc; close all;

% call of the function
myVal=5;
[f, g, h] = myfunction(myVal);
fprintf('f = %d     g = %d    h = %d', f, g, h);


% definition of the function
function [f, g, h] = myfunction(y)
    syms x
    q = exp(2*sin(x)) - x;
    f = double( vpa( subs( q, y) ) );
    dq = diff(q);
    g = double( vpa(subs( dq , y)) );
    dq2 = diff(dq);
    h = double( vpa(subs( dq2, y)) );
end

A much more usuful version , is the following. Where you can pass your mathematical function as an argument:

clearvars; clc; close all;

% define your mathematical function
syms x
myF=exp(2*sin(x)) - x;

% define the evaluation point
myVal=5;

% call your function calculator
[f, g, h] = myFunctionCalculator(myVal, myF, x);

% print results
fprintf('f = %d     g = %d    h = %d', f, g, h);


function [f, g, h] = myFunctionCalculator(xo, q, x)

    f = double( vpa( subs( q, x, xo) ) );
    dq = diff(q);
    g = double( vpa(subs( dq , x,  xo)) );
    dq2 = diff(dq);
    h = double( vpa(subs( dq2, x, xo)) );

end

.