#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)#
