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
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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 $\displaystyle 2\sin(x) \cos(x) = \sin(2x)$. $\displaystyle y' = 6\sin(x) \cos(x) (1+ \sin^2 x)^2$ is equally correct.
Originally Posted by e^(i*pi) Note that $\displaystyle 2\sin(x) \cos(x) = \sin(2x)$. $\displaystyle y' = 6\sin(x) \cos(x) (1+ \sin^2 x)^2$ is equally correct. Oh yes of course! Thank you very much, I can't believe I didn't spot that.
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