# Differentiation with trigonometry

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• April 29th 2010, 02:02 PM
CSG18
Differentiation with trigonometry
Can someone please show me how this happens?

y = (1+sin^2 x)^3

y'= 3sin2x (1+sin^2 x)^2

because I thought that the differentiated form of sin^2x = 2sinxcosx
• April 29th 2010, 02:08 PM
e^(i*pi)
Quote:

Originally Posted by CSG18
Can someone please show me how this happens?

y = (1+sin^2 x)^3

y'= 3sin2x (1+sin^2 x)^2

because I thought that the differentiated form of sin^2x = 2sinxcosx

Note that $2\sin(x) \cos(x) = \sin(2x)$.

$y' = 6\sin(x) \cos(x) (1+ \sin^2 x)^2$ is equally correct.
• April 29th 2010, 02:15 PM
CSG18
Quote:

Originally Posted by e^(i*pi)
Note that $2\sin(x) \cos(x) = \sin(2x)$.

$y' = 6\sin(x) \cos(x) (1+ \sin^2 x)^2$ is equally correct.

Oh yes of course! Thank you very much, I can't believe I didn't spot that.