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Derivative ^x, is this correct?

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]y=\sqrt{x} * 19^x[/tex]


    2. Relevant equations



    3. The attempt at a solution
    [tex]y=x^{1/2}*19^x[/tex]
    [tex]y'=1/2x^{-1/2}*19[/tex]
     
  2. jcsd
  3. Oct 25, 2009 #2
    You need to use the product rule. If you didn't know:
    [tex]19^x = e^{\ln 19^x} = e^{x\ln 19} = e^{(\ln 19)x}[/tex]

    I'm double-checking everything now. :redface:
     
  4. Oct 26, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, that's not right and I don't see how you could have got that! The derivative of a product is typically a sum of derivatives and the dervative of "[itex]19^x[/itex]" is definitely not 19.

    I would just take the logarithm of both sides immediately:
    [tex]ln(y)= ln(\sqrt{x}19^x)= ln(x^{1/2})+ ln(19^x)[/tex]
    [tex]ln(y)= (1/2)ln(x)+ x ln(19)[/tex]

    Now differentiate both sides and solve for y'.
     
  5. Oct 26, 2009 #4
    y = sqrt(x) * 10^x

    y` = sqrt(x) * (10^x)` + 10^x * (sqrt(x)) `
     
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