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. I'm guessing I shouldn't combine cos(x)cos(x)^2 though because I probably need that cos(x)^2 to use the identity.

Answer 1

:O

Answer 2

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Answer 3

I used product rule, but got a different answer than you, which simplifies to the answer you stated. Try looking at that calculation again.

Answer 4

Ok ok so I think I found a way to get the answer, I don't know if it's right or not, but yeah.
cos'(x)^2sin(x)+cos(x)^2sin'(x)
2cos(x)[-sin(x)]sin(x)+cos(x)^2cos(x)
cos(x)2[-sin(x)]sin(x)+1-sin(x)^2
cos(x)[-2sin(x)^2]+1-sin(x)^2
cos(x)[-3sin(x)^2+1)
cos(x)(1-3sin(x)^2)

Answer 5

I think that is basically what I did. I don't see anything incorrect. Good job.:)