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Mathematics 11 Online
OpenStudy (lucaz):

derivative of (sin(2x))^3? I need the steps, I think my answer is wrong

OpenStudy (amistre64):

chain rule is useful in this ... whats your answer?

OpenStudy (lucaz):

6sin^2(2x)cos(2x)

OpenStudy (lucaz):

maybe I'm doing the chain rule wrong?

OpenStudy (amistre64):

(sin(2x))^3 f(g(h)) f^3 = 3f^2 f', but f=sin(g) f' = cos(g) g', but g=2x g' = 2 3sin^2(2x) * cos(2x) *2 is right, but can prolly be simplified by some archaic trig rules

OpenStudy (amistre64):

s^2 = 1-c^2 6 (1-cos^2(2x)) * cos(2x) prolly not that route ... sin2x cos2x = 2sin(x/2) maybe?

OpenStudy (amistre64):

all depends on whos grading it.

OpenStudy (lucaz):

what I have here is 3sin(2x)sin(4x)

OpenStudy (amistre64):

that works too

OpenStudy (amistre64):

i had my sin2x cos2x a little off in my head :)

OpenStudy (lucaz):

ok then, thank you

OpenStudy (amistre64):

youre welcome

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