How to find the derivative of the function at some value of x?
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    reema shrestha
 le 1 Juil 2017
  
    
    
    
    
    Commenté : Walter Roberson
      
      
 le 24 Nov 2020
            I am trying to write a code for the Newton-Raphson iteration method. So at first i created a function
function y=f1(x)
y=x^2-3*x+2;
Then in next M-file,i wrote
 syms x;
 x(1)=0;
 i=1;
 x(i);
 z=f1(x(i))
But then I try to differentiate the function
 df= eval((subs(diff(f1,x,x(i)),x,x(i))))
it shows error.
how do i solve the problem? I tried various ways but none of them worked. How do i call the function from previous M-file?
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Réponse acceptée
  John D'Errico
      
      
 le 1 Juil 2017
        MATLAB cannot do symbolic differentiation on an m-file. That would in general be impossible, since you could stick anything you wanted in there.
You have two choices:
1. Perform the differentiation in advance, using the sym tools. Then you can pass the derivative function also to your NR code.
2. Inside the NR code, use finite differencing to compute an approximation to the derivative. This is almost always adequate for Newton schemes, although care must be taken to get a good estimate, using an appropriate step size. Also, central differences are considerable more accurate, so use them whenever possible.
2 commentaires
  Walter Roberson
      
      
 le 1 Juil 2017
				Consider the lines
a = 5;
b = a + 2;
a = 11;
Then what is the value of b afterwards? Is it now 13, because a is 11 and b = a + 2? Or is it 7 because the value of a at the time of the operation was looked up and used?
Likewise when you use
a = sym('a');
b = a + 2;
a = 11;
then what is the value of b? Is it now 11, because a is 11 and b = a + 2? Or is it sym('a')+2 because the value of a at the time of the operation was looked up and used?
If you were to use
a = sym('a');
b = a + 2;
a = 11;
subs(b, a, 5)
then b is sym('a')+2 and a has become (numeric) 11, and you are then asking to do
subs(b, 11, 5)
which does not change the sym('a') inside b.
When you assigned numeric 11 to a then it lost its identity as sym('a').
Moral of the story:
Never assign a new value to something that was previously a sym and expect anything to have been updated with the new value. subs() the new value in for the symbol instead:
 subs(g, x, 0)
Plus de réponses (2)
  Karan Gill
    
 le 5 Juil 2017
        
      Modifié(e) : Karan Gill
    
 le 17 Oct 2017
  
      To differentiate a function and then find the value, use symbolic functions. For details, see https://www.mathworks.com/help/symbolic/create-symbolic-functions.html
>> syms f(x)
>> f(x) = x^2 -3*x + 2
f(x) =
x^2 - 3*x + 2
>> g = diff(f)
g(x) =
2*x - 3
>> g(2) % value at x = 2
ans =
1
>> xValues = [-10 5 88]
xValues =
   -10     5    88
>> g(xValues)
ans =
[ -23, 7, 173]
3 commentaires
  Sarah Johnson
 le 12 Fév 2020
				This does not print out the values at g, when I plug it in it prints out the functions themselves
  Walter Roberson
      
      
 le 13 Fév 2020
				>> f(x) = x^2 -3*x + 2
f(x) =
x^2 - 3*x + 2
>> g = diff(f)
g(x) =
2*x - 3
>> g(7)
ans =
11
Works for me.
When I was glancing at your other recent post, it looked to me as if you have a different question: namely to determine the value of a constant in the formula such that the known boundary value was satisfied.
  Hamza saeed khan
 le 24 Nov 2020
        Undefined function or variable 'syms'.
Error in difff (line 2)
syms y(x)
1 commentaire
  Walter Roberson
      
      
 le 24 Nov 2020
				If you get that answer, then you do not have the Symbolic Toolbox installed and licensed.
There are some third-party symbolic packages that can be used with MATLAB with varying degrees of difficulty. Some of them are commercial; some of them are free (such as symbolic python).
If you do not have a symbolic software package of some kind, then derivatives can be calculated for some restricted cases such as polynomials or piecewise polynomials (including cubic spline); beyond that you start having to do numeric approximations of derivatives.
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