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How to find the derivative of this?

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Find dy/dx of this:
    Assume p>0, y=(p^(x^p))(x^p)

    2. Relevant equations

    1. d/dx f(x)g(x)h(x) = f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)


    3. The attempt at a solution

    ln(y)= p(ln(p^x))x^p

    1/y y'=p'(ln(p^x))x^p + p(logp)x^p + p(ln(p^x))px^(p-1) --use "revlevent" equation

    1/y y'= 0 + plogp(x^p) + p(ln(p^x))px^(p-1) -- simplify derivatives

    y'= y(plogp(x^p) + p(ln(p^x))px^(p-1)) -- multiply both sides by y

    y'= (p^(x^p))(x^p)(plogp(x^p) + p(ln(p^x))px^(p-1)) -- sub in y from beginning




    Can someone point out where my error is, and how to fix it?

    Thanks
     
  2. jcsd
  3. Oct 15, 2012 #2

    SammyS

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    Hello, ab94. Welcome to PF !

    Is this [itex]\displaystyle y=(p^{x^p})(x^p)\ ?[/itex]

    Then you have ln(y) wrong.
     
    Last edited: Oct 15, 2012
  4. Oct 15, 2012 #3
    Yes, that's what it is. So I can't take ln of both sides and say

    ln(y)=ln(p^x^p)
    ln(y)= pln(p^x) ?

    I thought We could bring down exponents when using that ln property.
     
  5. Oct 15, 2012 #4

    SammyS

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    No.

    [itex]\displaystyle \ln(y)=\ln\left((p^{x^p})(x^p)\right)[/itex]

    Now use some properties of logs & differentiate.
     
  6. Oct 15, 2012 #5
    so then would I split up the ln's into

    ln(y)=ln(p^x^p) + ln(x^p) ?

    If so I ended with y' = p^x^p(x^p) (plogp + p/x) ...don't think that's right
     
  7. Oct 15, 2012 #6

    SammyS

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    Show your steps, please.
     
  8. Oct 15, 2012 #7
    You almost got it.

    How did you expand ##ln(p^{x^{p}})##

    After this, take the derivative.

    A little late edit.
     
  9. Oct 15, 2012 #8
    Ohh thanks, I was taking the highest exponent (p) instead of the entire exponent (xp). I'll post my solution after I work it out
     
    Last edited: Oct 15, 2012
  10. Oct 15, 2012 #9
    Good.
     
  11. Oct 15, 2012 #10
    Ok so:

    ln(y)= ln(p^x^p) + ln(x^p)
    ln(y)=(x^p)ln(p) + (p)ln(x)
    1/y y'= (px^(p-1))ln(p) + (x^p)/p + 0 + p/x

    therfore y'= p^x^p(x^p)(px^(p-1))ln(p) + (x^p)/p + + p/x

    Is this correct? or is there an error
     
  12. Oct 15, 2012 #11
    You have the general idea, but your grouping is wrong.

    $$\frac{dy}{dx} ~=~ y\cdot [px^{p-1}ln(x)+\frac{x^{p}}{p}+\frac{p}{x}]$$
     
  13. Oct 15, 2012 #12
    Nvm I got the right answer, I used wolframalpha and I subtracted my answer from theirs, and got 0. (i fixed my own error up there ^)

    Thanks for the help
     
  14. Oct 15, 2012 #13

    SammyS

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    By the way: [itex]\displaystyle p=e^{\ln(p)}[/itex]

    Therefore,

    [itex]\displaystyle y=(p^{x^p})(x^p)[/itex]
    [itex]\displaystyle
    =e^{\ln(p)x^p}(x^p)[/itex]

    Taking the derivative:

    [itex]\displaystyle y'=\ln(p)\,p\,x^{p-1}e^{\ln(p)x^p}(x^p)+e^{\ln(p)x^p}px^{p-1}[/itex]

    For this problem it looks like a better way than doing the logarithmic derivative.
     
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