Find the value of x when the first derivative is equal to 0

Hello community,

I have the data x, y (.mat version) from which I compute its first derivative as follows :

dy = diff(y)./diff(x);

I need to find the points (x, y) such that the first derivative is 0. (In this case, there should be 2 points)

I have tried with this code, but it doesn't work, it tells me "Second argument must be a vector of symbolic variables"

solve(dy==0,x) 

Any idea how to fix the problem?

Thanks in advance!

Answer 1

[Torsten]

x = -pi/2:0.01:pi/2;
y = cos(x);
dydx = diff(y)./diff(x);
[~,idx] = min(dydx.^2);
xmin = (x(idx+1)+x(idx))/2

xmin is the point with slope closest to 0.

[Angelavtc]

Thank you very much @Torsten, but I don't understand your answer :( Why do you square the derivative? Then, according to my data, I would have to find two points such that the derivative is equal to 0. With your code, I only get one. Finally, how can I get the y in my data that belongs to the closest xmin points? Thank you again!

[Torsten]

function main
  x = -3*pi/2:0.01:3*pi/2;
  y = cos(x);
  dydx = diff(y)./diff(x);
  [B,I] = sort(dydx.^2);
  for i=1:3
    solx(i) = (x(I(i)) + x(I(i)+1))/2;
    soly(i) = (y(I(i)) + y(I(i)+1))/2;
  end
  solx(1),soly(1)
  solx(2),soly(2)
  solx(3),soly(3)
end

gives you the three points in [-3/2*pi:3/2*pi] where cos(x) has least, second least and third least slope.

Instead of squaring dydx, you could also have used abs(dydx). Can you imagine why this is necessary ?

[Angelavtc]

@Torsten I understand. With the squared derivative, the negative values become positive and the minimum is 0, the value we are looking for. Thank you very much :)