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Differentiate the following equation

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Differentiate

    s = Tan [itex]^{2}[/itex](e[itex]^{4t}[/itex])



    2. Relevant equations

    [e[itex]^{u}[/itex]]' = u' e[itex]^{u}[/itex]

    [Tan(θ)]' = Sec[itex]^{2}[/itex](θ)

    Sec(θ) = [itex]\frac{1}{Cos(θ)}[/itex]



    3. The attempt at a solution

    s' = 2[Tan(e[itex]^{4t}[/itex])] * Sec[itex]^{2}[/itex](e[itex]^{4t}[/itex]) * 4e[itex]^{4t}[/itex]

    s' = 8e[itex]^{4t}[/itex] * Tan(e[itex]^{4t}[/itex]) * Sec[itex]^{2}[/itex](e[itex]^{4t}[/itex])

    s' = 8e[itex]^{4t}[/itex] * [itex]\frac{Tan(e^(4t))}{Cos^2(e^(4t))}[/itex]



    Mathway says that s' = 0... Can someone please let me know where I went wrong? Thanks
     
  2. jcsd
  3. Oct 16, 2012 #2
    You're derivative is right. Are you sure they didn't ask you to evaluate the derivative at some point? I'm guessing that they want you to evaluate it at 0.
     
  4. Oct 17, 2012 #3
    Yes, you're right, I have to find s'(0) aka v(0)

    I end up getting this:

    v(0) = 8e[itex]^{0}[/itex] * [itex]\frac{Tan(e^0)}{Cos^2(e^0)}[/itex]

    v(0) = 8 * [itex]\frac{Tan(1)}{Cos^2(1)}[/itex]

    Is that good?
     
  5. Oct 21, 2012 #4
    Yes, it is.
     
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