I assume that you know how to use the chain rule and the product rule. If you don't please ask for further explanation in the comments.
Define: f(x)= ln[x^4sin^2(x)]f(x)=ln[x4sin2(x)]
First use the chain rule:
f'(x)=(1/(x^4sin^2(x))) (d/dx(x^4sin^2(x)))
Use the product rule to find d/dx(x^4sin^2(x))
To do this we'll need to know d/dx sin^2(x):
d/dx sin^2(x)= d/dx sin(x)sin(x)
Use the product rule:
d/dx sin(x)sin(x)=sin(x)cos(x)+cos(x)sin(x)=2sin(x)cos(x)
Using this result:
f'(x)=(1/(x^4sin^2(x))) (4x^3sin^2(x)+x^4(2sin(x)cos(x)))
=(1/(x^4sin^2(x))) (4x^3)(sin^2(x)+x/2 sin(x)cos(x))
=(4/(xsin^2(x))) (sin^2(x)+x/2 sin(x)cos(x))
=(4/(x)) (sin^2(x)/sin^2(x)+x/2(sin(x)cos(x))/(sin^2(x)) )
=(4/(x)) (sin^2(x)/sin^2(x)+x/2 (sin(x)cos(x))/(sin(x)sin(x)))
=(4/(x)) (1+x/2 cos(x)/sin(x)))
Note that, cot(x)=1/tan(x)=cos(x)/sin(x), so:
f'(x)=(4/(x)) (1+x/2 cot(x))
Finally we have:
d/dx ln[x^4sin^2(x)]=4/(x)+2 cot(x)
