How to find derivative equal to 0?

So I'm trying to find where the derivative of a function is equal to zero, and at what x-value.

x8 = linspace(-2, 2, 100);

y8diff = -(((x8.^2 - 1).*sin(x8 + 0.1) + x8.*cos(x8 + 0.1))./((sqrt(1-x8.^2).*(cos(x8 + 0.1).^2))));

We're not allowed to use MatLabs Toolbox

[Walter Roberson]

Is y8diff the function or is it already the derivative?

[Amanda]

Oh sorry for not clarifying, it is the derivative

[Star Strider]

What are you allowed to use?

x8 = linspace(-2, 2, 100);
y8diff = -(((x8.^2 - 1).*sin(x8 + 0.1) + x8.*cos(x8 + 0.1))./((sqrt(1-x8.^2).*(cos(x8 + 0.1).^2))));

figure
yyaxis left
plot(x8, real(y8diff));
ylabel('Real')
yyaxis right
plot(x8, imag(y8diff))
ylabel('Imaginary')
grid

.

[Amanda]

I guess everything except for syms x

Answer 1

[Walter Roberson]

y8diff = @(x8) -(((x8.^2 - 1).*sin(x8 + 0.1) + x8.*cos(x8 + 0.1))./((sqrt(1-x8.^2).*(cos(x8 + 0.1).^2))));

That is, turn it into an anonymous function. Now create a second anonymous function that is the square of the above, and fzero the second function.

The graph that Star Strider created hints that there might be two zeros, both close to 1.5

[Amanda]

what should my second argument in fzero be?

[Walter Roberson]

the initial guess. You should probably try with a guess below 1.4 and another above 1.6

[Walter Roberson]

y8diff = @(x8) -(((x8.^2 - 1).*sin(x8 + 0.1) + x8.*cos(x8 + 0.1))./((sqrt(1-x8.^2).*(cos(x8 + 0.1).^2))));
syms x8
y8d = y8diff(x8);
eqn = y8d.^4

eqn = 

vpasolve(eqn, 1.4)

ans = 

0.52134986062092355768691638558689

vpasolve(eqn, 1.5)

ans = 

2.6217707005409009447216985496563

fplot(eqn, [0.5 1.2])

fplot([real(y8d), imag(y8d)], [0 0.8])

fplot([real(y8d), imag(y8d)], [2 2.7])

X0 = linspace(0,40, 100);
roots_found = arrayfun(@(X) double(vpasolve(eqn, X)), X0)

roots_found = 1×100

0.5213 0.5213 0.5213 2.6218 2.6218 2.6218 2.6218 2.6218 2.6218 2.6218 2.6218 2.6218 6.0138 6.0138 6.0138 6.0138 6.0138 6.0138 6.0138 6.0138 9.2154 9.2154 9.2154 9.2154 9.2154 9.2154 9.2154 12.3853 12.3853 12.3853

plot(X0, roots_found)

u = uniquetol(roots_found)

u = 1×14

0.5213 2.6218 6.0138 9.2154 12.3853 15.5434 18.6960 21.8453 24.9927 28.1388 31.2839 34.4285 37.5725 40.7161

diff(u)

ans = 1×13

2.1004 3.3920 3.2016 3.1699 3.1582 3.1525 3.1493 3.1474 3.1461 3.1452 3.1445 3.1440 3.1437

There seem to be an indefinite number of roots with the difference between them looking like it tends to pi

Answer 2

[Image Analyst]

So you want where the real part of y8diff and the imaginary part of y8diff are both zero at the same time. Like the two places around 1.4 and 1.5 in Star's plot above? I think you'll have to do it symbolically since if you do it numerically, there is no index where both the real and imaginary parts are zero.

%Star's code but with a lot finer sampling.
x8 = linspace(-2, 2, 10000);
y8diff = -(((x8.^2 - 1).*sin(x8 + 0.1) + x8.*cos(x8 + 0.1))./((sqrt(1-x8.^2).*(cos(x8 + 0.1).^2))));
figure
yyaxis left
plot(x8, real(y8diff));
ylabel('Real')
yyaxis right
plot(x8, imag(y8diff))
ylabel('Imaginary')
grid

% Now see where real and imaginary parts are both zero at the same time:
yr = real(y8diff);
yi = imag(y8diff);
index0 = find((yr == 0) & (yi == 0))

index0 =

  1×0 empty double row vector

[Amanda]

I'm only looking for real solutions. When I try to use find(real(y8diff) == 0), I get an error message saying:

"Check for incorrect argument data type or missing argument in call to function 'real'."

Edit: Nevermind, solved the error message. However, I get an 1x50 double vector. What should I do with it?

[Image Analyst]

That tells you that there are 50 elements in your vector that have a value of exactly 0. I don't know what you plan on doing with that information -- you didn't tell us. You said you just have to find out where it is zero.