How do you differentiate #y = sin^2(x) + sinx#?

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Answer 1 · 2016-09-04T00:22:58

#y' = sin2x + cosx#

#y = sinx(sinx + 1)#

#y' = cosx(sinx + 1) + sinx(cosx)#

#y' = cosxsinx + cosx + sinxcosx#

#y' = 2sinxcosx + cosx#

#y' = sin2x + cosx#

Hopefully this helps!