numerical derivative in matlab

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Christoppe
Christoppe 2022-8-4
评论： Star Strider 2024-1-5
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)
  2 个评论
Torsten
Torsten 2022-8-4
dy = subs(g,x,x0)
if you rename x in x0 in the preceeding input-command for x.
Dyuman Joshi
Dyuman Joshi 2024-1-5
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?

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回答（1 个）

Ayush
Ayush 2024-1-5
编辑：Ayush 2024-1-5
Ran in:
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:
Regards,
Ayush
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Dyuman Joshi
Dyuman Joshi 2024-1-5
I see. Thank you for your response.

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