numerical derivative in matlab

Hi! How can I show the numerical value of the first derivative of a function in this code?

(-0.1*x^4) - (0.15*x^3) - (0.5*x^2) - (0.25*x) + 1.2

Enter Step Size:

0.5

Enter the point you want to approximate the Derivative:

0.5

clear
clc
format short

syms x
P=input ('Enter the Equation: ','s');
f=inline(P)
g=diff(f(x),x);

h=input ('Enter Step Size: ')

x=input ('Enter the point you want to approximate the Derivative: ')
fprintf('f(x) = %0.4f', f(x))

dy=g(x) % How Can I get the derivative equation of g in numerical answer by substituting the value of x?

fprintf('x(xi+1) = %0.4f', x+(1*h))
fprintf('f(xi+1) = %0.4f', f(x+(1*h)))

FDD=(f(x+(1*h))-f(x))/h;
fprintf('FDD (Truncated) = %0.4f', (f(x+(1*h))-f(x))/h)

[Torsten]

dy = subs(g,x,x0)

if you rename x in x0 in the preceeding input-command for x.

[Dyuman Joshi]

inline is not recommended.

Simply define the expression/function by typing or use str2sym.

Don't you think the step size is too big for calculating numerical derivative?

Answer 1

[Ayush]

Hi Christoppe,

I understand that you want to show the numerical value of the first derivative of a function in the given code.

To do so, you need to substitute the value of “x” into the derivative function “g” and then evaluate it. This can be done by using the “subs” function or by converting the symbolic expression to a function and then evaluating it at the given point.

Refer the modified code below for better understanding:

clear
clc
format short 
syms x 
% P = input('Enter the Equation: ','s'); 
P = (-0.1*x^4) - (0.15*x^3) - (0.5*x^2) - (0.25*x) + 1.2;
f = inline(P); 
g = diff(f(x),x); 
% h = input('Enter Step Size: ');
h = 0.5;
% x_val = input('Enter the point you want to approximate the Derivative: '); 
x_val = 0.5;
fprintf('f(x) = %0.4f\n', f(x_val)); % Display the value of the function at x 

f(x) = 0.9250

% Calculate the numerical value of the derivative at x 
g_val = double(subs(g, x, x_val)); 
fprintf('g(x) = %0.4f\n', g_val); % Display the numerical value of the derivative 

g(x) = -0.9125

fprintf('x(xi+1) = %0.4f\n', x_val + h); 

x(xi+1) = 1.0000

fprintf('f(xi+1) = %0.4f\n', f(x_val + h)); 

f(xi+1) = 0.2000

FDD = (f(x_val + h) - f(x_val)) / h; 
fprintf('FDD (Truncated) = %0.4f\n', FDD);

FDD (Truncated) = -1.4500

The "inline" function is not recommended, instead use the "anonymous function". For more information on the “subs” function and "anonymous function" refer the documentation page given below:

• "subs function": https://www.mathworks.com/help/symbolic/subs.html
• "anonymous function": https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html

Regards,

Ayush

[Stephen23]

Two questions:

1. Why do you use deprecated INLINE ?
2. Are you TMW staff?

[Dyuman Joshi]

Stephen, I suspect that Ayush is a TMW staff member, as I have seen a staff account named Ayush upvoting the answers from this account (in succession) awhile ago.

They might have been an intern at TMW.

[Star Strider]

I saw that as well, and reported it to the appropriate person at TMW.

[Dyuman Joshi]

@Star Strider, Have you gotten any response/update from TMW about it?

[Star Strider]

Not yet, other than the person promising to look into it. (I provided the dates and times.) I reported it when I saw it, so a few minutes after it occurred.

[Dyuman Joshi]

I see. Thank you for your response.

[Star Strider]

My pleasure!