Heres what I said
2sinxcosx=sin2x isn't that a rule?
then y=sin2x
then the derivative of that is
2cos2x
right
or do u have to do it as a product?
2007-11-01 06:24:22 · 5 answers · asked by limallama 4 in Science & Mathematics ➔ Mathematics
What youdid is correct but you can also do it as a product and should come up with the same answer.
y = 2sinxcosx
y' = 2cosx(cosx) + 2sinx(-sinx)
y' = 2cos^2x - 2sin^2x
y' = 2(cos^2x - sin^2x)
But cos2x = cos^2x - sin^2x
y' = 2cos2x
So you get the same result either way. Clearly though it is easier to do it the way you did.
2007-11-01 06:29:31 · answer #1 · answered by Astral Walker 7 · 1⤊ 0⤋
Derivative Of 2sinxcosx
2016-11-14 07:20:13 · answer #2 · answered by ? 4 · 0⤊ 0⤋
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RE:
What is the derivative of Y =2sinxcosx?
Heres what I said
2sinxcosx=sin2x isn't that a rule?
then y=sin2x
then the derivative of that is
2cos2x
right
or do u have to do it as a product?
2015-08-14 07:04:13 · answer #3 · answered by ? 1 · 0⤊ 0⤋
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This equals sin2x. y' = 2cos2x
2016-04-01 02:17:51 · answer #4 · answered by Anonymous · 0⤊ 0⤋
2 cosx (-sinx) = -2cosxsinx y' secx= secxtanx
2016-03-18 11:45:58 · answer #5 · answered by Sylvia 4 · 0⤊ 0⤋
you have the correct solution... you use the chain rule which was correct
2007-11-01 06:29:07 · answer #6 · answered by Anonymous · 0⤊ 0⤋
y = sin (2x)
let u = 2x
du / dx = 2
y = sin u
dy/du = cos u
dy/dx = (du/dx)(dy/du)
dy/dx = 2 cos u
dy/dx = 2 cos 2x
Agrees with your answer.
2007-11-01 08:27:03 · answer #7 · answered by Como 7 · 2⤊ 0⤋