HELLO!
I'm new here and i sorta need help in this stupid dy/dx thing. Not really need that much help, but I just wanna confirm whether if I got it right (finally?)
dy/dx of ln(sin²2x) = cot2x?
Thanks for your help in advance!
Hi,
It is easier to learn the general result, i.e. d/dx [ ln f(x) ] = f'(x) / f(x).
So d/dx ln(sin²2x) = {2 (sin 2x) (cos 2x) (2)} / sin²2x. Thanks!
Cheers,
Wen Shih
Oh haha. I had 4cot2x in my workings and I kinda left it out cos I was scribbling. So i am right? 4cot2x?