Question

derivative of Sec^2(3x) anyone? totally lost

Answer 1

\[\sec^2(3x) dx = { 6\sin(3x) \over (\cos(3x)^3)} \]

Answer 2

Remember the Chain Rule? Think of the problem as y = (sec3x)^2. Then dy/dx will be 2sec3x times d/dx(sec3x). Now all you need to do is differentiate sec3x. Hint: think of sec3w as 1/cos3x and use the quotient rule.

Answer 3

Thank you, that really helped clear the air!