cscx = 1/sinxcscx=1sinx
so
-csc^2x = -1/sin^2x−csc2x=−1sin2x
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)
