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Derivative of a trig function/chain rule

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Differentiate [tex]sin^3 2x[/tex]


    2. Relevant equations

    Chain rule.

    3. The attempt at a solution

    I just need to see if I got the answer right, there aren't any solutions available for this paper, and I need it for the next parts of a question...

    I got...
    [tex]
    \begin{align*}
    = (sin 2x)^3\\
    differentiate\\
    = 3 (sin 2x)^2 (2 cos 2x)\\
    = 6 cos 2x sin^2 2x\\
    \end{align*}
    [/tex]

    ?
     
  2. jcsd
  3. Apr 8, 2009 #2

    dx

    User Avatar
    Homework Helper
    Gold Member

    Correct.
     
  4. Apr 8, 2009 #3
    Sweet!
    Muchos gracias!
     
  5. Apr 8, 2009 #4
    Another quickie...

    Is the derivative of [tex]y=xe^{-3x}[/tex] going to work out to be [tex]e^{-3x} (1-3x)[/tex] ?
     
  6. Apr 8, 2009 #5

    Mark44

    Staff: Mentor

    Yes.

    Also, that would be muchas gracias.
     
  7. Apr 8, 2009 #6

    Mark44

    Staff: Mentor

    Some help with your notation.

    d/dx(sin3(2x))
    = 3 (sin 2x)^2 (2 cos 2x)
    = 6 cos(2x) sin2(2x)

    The first line says that we want to differentiate the expression in parentheses. In the second line, we have taken the derivative, using the chain rule. The third line is still the derivative, but in cleaned-up form.
     
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