Asked by Tom
Find f'(x) for f(x) = sin3(3x2).
Answers
Answered by
Tom
Sorry it is sin^3(3x^2)
Answered by
Reiny
look at it as:
f(x) = (sin(3x^2))^3
f ' (x) = 3(sin(3x^2))^2 cos(3x^2) (6x)
= 18x sin^2 (3x^2) cos(3x^2)
or
= 9x sin(6x^2) cos(3x^2), using property sin 2A = 2sinAcosA
f(x) = (sin(3x^2))^3
f ' (x) = 3(sin(3x^2))^2 cos(3x^2) (6x)
= 18x sin^2 (3x^2) cos(3x^2)
or
= 9x sin(6x^2) cos(3x^2), using property sin 2A = 2sinAcosA
Answered by
Steve
Hmm. I see an extra sin, not cos:
9x sin(6x^2) sin(3x^2)
9x sin(6x^2) sin(3x^2)
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