Find the derivative of [cos(x)^2][sin(x)] and it has to equal cos(x)[1-3sin(x)^2]. Also you have to use the identity sin(x)^2+cos(x)^2=1.
So far I have:
Using product rule- u'(x)v(x)+v'(x)u(x)
=[(cos'(x)^2)*sin(x)] + [sin'(x)*cos(x)^2]
using chain rule
=[2cos(x)*cos'(x)*sin(x)] + [cos(x)*cos(x)^2]
simplifying
=2cos(x)[sin(x)^2]+cos(x)^3
And after this I don't know what to do, but I may have messed up on a previous step or have overlooked an identity I could have used.

Question

Find the derivative of [cos(x)^2][sin(x)] and it has to equal cos(x)[1-3sin(x)^2].  Also you have to use the identity sin(x)^2+cos(x)^2=1.
So far I have:
Using product rule- u'(x)v(x)+v'(x)u(x)
=[(cos'(x)^2)*sin(x)] + [sin'(x)*cos(x)^2]
using chain rule 
=[2cos(x)*cos'(x)*sin(x)] + [cos(x)*cos(x)^2]
simplifying
=2cos(x)[sin(x)^2]+cos(x)^3
And after this I don't know what to do, but I may have messed up on a previous step or have overlooked an identity I could have used.

anonymous · Answer

use the identity you posted sin(x)^2+cos(x)^2=1 and rearrange for cos(x)^2

cos(x)^2 = 1-sin(x)^2

plug it in:

[sin(x)]1-sin(x)^2 = sin(x) - sin(x)^3

take the derivative:

cos(x) -3sin(x)^2 <---- answer

anonymous · Answer

I don't understand how you got sin(x)^3. Also I kind of learned it that you set it up in product/quotient/etc before you do any identities. Finally, when you took the derivative, it needs to be [1-3sin(x)^2] and so I think a one got dropped off at some point.

goformit100 · Answer

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anonymous · Answer

Do you know the distributive property? @mmorrison52