Question

How do you find the derivative of `sin^2(3x)/cos(2x)`?

Answer 1

`f'(x)=frac(6sin(3x)cos(3x)cos(2x)+2sin^2(3x)sin(2x))(cos^2(2x))`

Explanation:

Use the Quotient Rule.
`f'(x)=(color(red)(d/(dx)[sin^2(3x)])*cos(2x)-sin^2(3x)*color(blue)(d/(dx)[cos(2x)]))/cos^2(2x)`

Use the Chain Rule to find both derivatives inside the quotient rule.
`color(red)(d/(dx)[sin^2(3x)])=2sin(3x)*d/(dx)[sin(3x)]=2sin(3x)*cos(3x)*d/(dx)[3x]=color(red)(6sin(3x)cos(3x)`
`color(blue)(d/(dx)[cos(2x)])=-sin(2x)*d/(dx)[2x]=color(blue)(-2sin(2x)`

Plug back in.
`color(green)(f'(x)=frac(6sin(3x)cos(3x)cos(2x)+2sin^2(3x)sin(2x))(cos^2(2x)))`