Posted by Terry on Monday, May 5, 2008 at 8:19pm.
f(x) = x² + 2Cos²x, find f ' (x)
a) 2(x+cos x)
b) x  sin x
c) 2x + sin x
d)2(x  sin2x)
I got neither of these answers, since the 2nd part should be chain rule, right?
f(x) = x² + 2Cos²x = x² + 2(Cos x)²
then f '(x) = (2)(2)(cosx)(sinx)
My answer:
f '(x) = 2x  4cosxsinx
If I do it this way, then I get that the answer is d:
f(x) = x² + 2Cos²x
f '(x) = x² + 2(sin x)²
f '(x) = 2x  2sin²x
f '(x) = 2(x  sin²x)
Is that how I should be doing it?

Math, derivatives  Terry, Monday, May 5, 2008 at 8:20pm
sorry, this part:
then f '(x) = (2)(2)(cosx)(sinx)
should say:
then f '(x) = 2x + (2)(2)(cosx)(sinx)

Math, derivatives  Doug, Monday, May 5, 2008 at 8:57pm
You're on the right track.
sin 2x= 2sin x cos x (identity)
substitute into your first answer.
f'(x)=2x2(sin2x)
f'(x)=2(xsin2x)

Math, derivatives  Terry, Monday, May 5, 2008 at 9:24pm
Thanks for your reply! That makes much more sense now. I guess I should go and review my identities.
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