Find the derivative of y=10csc(3x/2) at x=2pi/6?

We'll recall the identity that gives the cosecant
function:

csc x = 1/sin x

y =
10csc(3x/2) = 10/sin(3x/2)

We'll calculate the 1st
derivative of the function using the quotient rule:

dy/dx =
[(10)'*sin(3x/2) - 10*[sin(3x/2)]'/[sin(3x/2)]^2

dy/dx = -
15*[cos(3x/2)]/[sin(3x/2)]^2

Now, we'll calculate the value
of the first derivative at x = 2 /6

dy/dx = -
15*[cos(6 /12)]/[sin(6 /12)]^2

dy/dx = -15*cos(
/2)/(sin /2)^2

But cos /2 = 0, therefore dy/dx =
0 at x = 2 /6.

Therefore the function has an extreme at
x = 2 /6.

The requested value of the 1st
derivative, at x = 2 /6 is dy/dx = 0.