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Posted by on Sunday, September 18, 2011 at 2:16pm.

y=sin^2(x)- cos^2(x)

I have this:
y'= 2cosx + 2sinx
What do i do next??

  • Calculus - , Sunday, September 18, 2011 at 5:01pm

    Ahhh, you might recognize that
    cos^2 x - sin^2 x = cos 2x

    so y=sin^2(x)- cos^2(x)
    = - cos 2x

    dy/dx = 2sin 2x or 4sinxcosx

    If you don't see that identity right away, then

    dy/dx = 2(sinx)cosx - 2cosx(-sinx)
    = 4sinxcosx or 2sin(2x)

  • Calculus - , Monday, September 19, 2011 at 10:39am

    Oh! Thank you so much! I completely forget about the double angle identity!!! :)

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