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

use the chain rule

`y' = 10 (d( csc((3x)/2)))/(d((3x)/2)) * (d((3x)/2)/2)/dx`

Since `csc(u) = 1/(sin(u))`

`(d(csc(u)))/(du) = -1/(sin^2(u))*(cos(u))=-cot(u)csc(u)`

So the answer is

`y'=-10cot((3x)/2)csc((3x)/2)*3/2=-15cot((3x)/2))csc((3x)/2)`

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`pi` /6

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

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

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

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

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