Hello!
I'm having trouble when i try to derivate this function:
$\displaystyle y=sin^6x+cos^6x+3sin^2xcos^2x$
Any idea?
Thanks.
P.s solution in my book is 0.
Hello !
This answer is effectively 0
You can derivate (awful job !)
Or you can see that
$\displaystyle (sin^2x+cos^2x)^3=sin^6x+3sin^4xcos^2x+3sin^2xcos^ 4x+cos^6x$
but
$\displaystyle sin^2x+cos^2x=1$
therefore
$\displaystyle 1=sin^6x+3sin^2xcos^2x(sin^2x+cos^2x)+cos^6x$
1=y and then y'=0