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.

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- Nov 29th 2008, 03:56 AMGreenMilehelp finding derivative
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. - Nov 29th 2008, 05:08 AMrunning-gag
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 - Nov 29th 2008, 09:59 AMGreenMile
Sorry how did u get $\displaystyle (sin^2x+cos^2x)^3=sin^6x+3sin^4xcos^2x+3sin^2xcos^ 4x+cos^6x$

- Nov 29th 2008, 10:21 AMskeeter
$\displaystyle (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$