the average rate of change of the function to find the derivative

I am trying to write a program to find f'(x)=f(x+h)-f(x)/h at different h

I could not get the result. I do not know where is my mistake could you please help me

Answer 1

[David Goodmanson]

Hello Asma

Take a look at the following

>> 1:8       ans =  1     2     3     4     5     6     7     8
>> -1:-8     ans = 1×0 empty double row vector
>> -1:-1:-8  ans =  -1    -2    -3    -4    -5    -6    -7    -8

Matlab assumes by default that indices are incremented by +1. If you want to decrement by -1 you have to supply that information as on the last line. That change gets the for loop going. However, you are not saving the answer for different values of k. Values can be saved in an array z, but indices that address an array must be positive. So it's better to have positive values of k and adjust accordingly.

z = zeros(1,8)  % make an initial z array
for k = 1:8 
  h = 10.^(-k) 
  z(k) = (exp(1+h) - exp(1))./h 
end
disp(z)

There is no need to define a new function for exp since you can just use exp directly.

[David Goodmanson]

Hi Asma,

I didn't get back on this quite soon enough. First, in place of disp(z) it's easier to see the results if z is made into a column vector

format long
disp(z')

I think your intention was to take a look at the difference between the approximate derivative and the exact derivative, which is exp(1). This is

format shorte
disp(z'-exp(1))

which does have a numerical problem at h = 1e-8.