What is the first derivative of sin³ (cosec t) with respect to t

My workings as follows:

d/dt sin³ (cosec t) = 3 sin ² (cosec t) x ( -1/sin²t x cos t)

= 3 cot t cosec t sin ² (cosec t)

Is this answer correct? Thanks

y = [ sin ( cosec t ) ] ³

let u = cosec t
du/dt = - cosec t cot t

y = [ sin u ] ³

dy/du = 3 sin ² u cos u

dy/dt = - 3 sin ² u cos u cosec t cot t

dy/dt = - 3 [ sin ² ( cosec t ) ] [ cos ( cosec t ) ] [ cosec t ] [ cot t ]

Let's look at this one piece at a time since we have to use the chain rule. Remember that sin^3(csc(t)) is the same as (sin(csc(t))^3.

d/dt(sin(csc(t))^3=3(sin(csc(t))^2(cos…
=(-3csc(t)(cot(t))(cos(csc(t))(sin(csc…

You are missing one thing. You also need to differentiate the sin^3. This will give you
-3cos(csc(t)) * cot(t) * csc(t) * sin^(2)(csc(t))

d/dt(sin^3(csc t)) = 3*sin²(csc t))*cos(csc t)*-csc(t) * cot(t)