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what is derivative of sin^2 (3x-1)^2 ?

dy/dx (sin^2 (x))= 2 sin x cos x
dy/dx (3x-1)^2= 6(3x-1)

chain rule?
= 2 sin ((3x-1)^2) cos ((3x-1)^2) * 6(3x-1)

Please help. Thank you

  • Derivatives -

    I'm not sure if this is correct. Can someone clarify thanks.

  • Derivatives -

    you are correct
    Or, you could go on and say

    sin (2(3x-1)^2) * 6(3x-1)

  • Derivatives -

    I got 12 sin (3x-1)^2 cos (3x-1)^2 * (3x-1) as my final answer. Is this in the correct simplified form ?

  • Derivatives -

    it is ok, but you can use

    2sinu cosu = sin(2u)

    to make it what I had:

    6(3x-1) sin(2(3x-1)^2)

    If you want to retain your original argument of sin and cos, then leave it as you had it.

  • Derivatives -

    I have taken that into consideration; thank you!

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