Matlab problem: How to calculate a rough, approximate, derivative vector by using the following formula?

Here is the complete task which I need to solve:

We know that the derivative is the coefficient k of a tangent line with mathematical expression, y=kx + m, that goes through a certain point, P, on a function curve. Lets start with a simple function, namely, y=x^2.

(a) Calculate and Plot this function for -10 < x < 10 with small enough step length to get a smooth curve.

(b) Now, In (a) you calculated two equally long vectors, one for the chosen x values and one with the calculated y values. From these two vectors, we can calculate a rough, approximate, derivative vector that we can call Yprimenum: Use a for-loop to calculate each element of Yprimenum according to the formula:

Yprimenum(i) = (y(i+1) – y(i)) / ∆x Where ∆x is the x step length, or equivallently x(i) – x(i-1)

NOTE: (a) is solved and for the (b) I wrote the following code but not working.

if true
   x=-10:10; 
for i=1: length(x); 
for y=x.^2 
Yprimenum=(y(i+1)-y(i))./(x(i)-x(i-1)); 
end; 
end; 
display Yprimenum;
end

[Star Strider]

And ‘not working’ means ... ?

[John D'Errico]

Why do you think it necessary to add the "if true" conditional around your code? Silly.

[Jabir Al Fatah]

ahh u should know that. if true was added automatically while pasting code here and i just forgot to modify that.

Answer 1

[George Papazafeiropoulos]

x=-10:10; 
y=x.^2;
Yprimenum=diff(y)./diff(x)

[Jabir Al Fatah]

i was just saying that my code does dont produce the expected result, rather it has an error. Thanks for editing, but i have to use for loops. i just solved it now like that:

for x=-10:10; for y=x.^2; Yprimenum=diff(y)./diff(x) end; end;

[Jabir Al Fatah]

hi again my solution just gives me some emply matrices

[George Papazafeiropoulos]

x=-10:10;
for I=2:length(x)
    y(I)=x(I)^2;
    Yprimenum(I-1)=(y(I)-y(I-1))/(x(I)-x(I-1))
end

[Jabir Al Fatah]

But here this formula "Yprimenum(I-1)=(y(I)-y(I-1))/(x(I)-x(I-1))" doesn't follow mine. And also why it's (I-1) with Yprimenum? Moreover, I need to plot Yprimenum in the same figure as Y and Yprime. What is the ∆X value ? I am eagerly waiting for your update of your answer PLEASE! Thank you.

[George Papazafeiropoulos]

Try this:

% 1st way: for loop
Yprimenum1=zeros(1,2*10);
x=-10:10;
y=x.^2;
for I=2:length(x)
    Yprimenum1(I-1)=(y(I)-y(I-1))/(x(I)-x(I-1));
end
% 2nd way: diff
Yprimenum2=diff(y)./diff(x);
% test the results
all(Yprimenum1==Yprimenum2)

Be careful that the length of the approximate derivative is equal to the length of the initial vectors minus 1