#cscx = 1/sinx#

so

#-csc^2x = -1/sin^2x#

Using the quotient rule, we know that

#d/dx f(x)/g(x) = (f'(x)g(x)-g'(x)f(x))/g^2(x)#

If we say that #f(x)=-1# and #g(x)=sin^2(x)#, then

#f'(x)=0#

#g'(x)=2sin(x)cos(x)#

and, putting these into the quotient rule formula,

#d/dx f(x)/g(x) = (--1*2sin(x)cos(x))/sin^4(x)#

#=(2cos(x))/sin^3(x)#

This gives the first derivative.

The second derivative is the derivative of the first derivative, so do the whole process again.

Use new #f(x) = 2cos(x)# and #g(x)=sin^3(x)#.

#f'(x)=-2sin(x)#

#g'(x)=d/dx(sin^2(x)*sin(x)) =#

#d/dxsin^2(x)*sin(x)+sin^2(x)*d/dxsin(x)=#

#2sin(x)cos(x)*sin(x)+sin^2(x)*cos(x)=3sin^2(x)cos(x)#

so

#g'(x)=3sin^2(x)cos(x)#

Now, using the formula for the quotient rule,

#d/dx f(x)/g(x) = (f'(x)g(x)-g'(x)f(x))/g^2(x)#

#=(-2sin^4(x)-6sin^2(x)cos^2(x))/sin^6(x)#

#=-2/sin^2(x)-(6cos^2(x))/sin^4(x)#

#=-2csc^2(x)-6cot^2(x)csc^2(x)#