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Determine dy/dx if y=(x^5)^lnx

  1. Aug 11, 2015 #1
    My attempt:

    lny=ln(x^5)^lnx
    lny=lnxlnx^5
    1/y.dy/dx=(lnx)d/dx(lnx^5)+d/dx(lnx)(lnx^5)
    = lnx(1/x^5)+1/x(lnx^5)
    = (lnx/x^5)+(lnx^5/x) <---- not sure if this simplifies to 1/x^4+lnx^4?
    dy/dx= y(lnx/x^5 + lnx^5/x)
    = (x^5)^lnx(lnx/x^5 + lnx^5/x)

    Any help would be appreciated.
    D.
     
  2. jcsd
  3. Aug 11, 2015 #2

    RUber

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    So far so good
    You forgot the chain rule. It should be:
    ##\ln x \frac{5x^4}{x^5} + \frac{\ln x^5 }{x}##
    This will simplify nicely.
     
  4. Aug 11, 2015 #3

    SteamKing

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    Strictly speaking, this is not a thread suitable for the Differential Equations forum.
     
  5. Aug 12, 2015 #4
    Thanks for your response RUber.
    ##\frac{d}{dx} lnx^5 = 5lnx^4##? Does it not? I am unsure how you got ##\ln x \frac{5x^4}{x^5}##
     
  6. Aug 12, 2015 #5
    It has been moved, my apologies.
     
  7. Aug 12, 2015 #6

    pwsnafu

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    No. ##\ln x^5 = 5 \ln x##.
     
  8. Aug 12, 2015 #7

    RUber

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    Let g(x) = x^5, then you have ##\frac{d}{dx} \ln g(x) = \frac{1}{g(x)} g'(x) ## by the chain rule.

    edit: Or you can take the simplification that pwsnafu gave first, and not have to worry about the chain rule.
     
  9. Aug 12, 2015 #8

    SteamKing

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    You can't ignore those two letters "ln" in front of x5. They mean something special.
     
  10. Aug 12, 2015 #9

    Mark44

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    Not (as already mentioned).
    ##\frac{d}{dx} lnx^5 = \frac{d}{dx} 5 lnx = 5 \frac{d}{dx} lnx## = ??
    I am assuming that lnx^5 means ln(x5) and not (ln(x))5.
    BTW, your original thread title was "Determine dy/dx of y=(x^5)^lnx". I changed "of" to "if" because you don't take "dy/dx" of something. This symbol already represents the derivative of y with respect to x. It is not a synonym of "take the derivative of".
     
  11. Aug 12, 2015 #10
    Okay let's see if I understand:

    Determine ##\frac{dy}{dx}## if y=##(x^5)^{lnx}##

    lny=ln##(x^5)^{lnx}##
    lny=##lnxlnx^5##
    ##\frac{1}{y}## ##\frac{dy}{dx}## = (lnx)##\frac{d}{dx}(lnx^5) +\frac{d}{dx}(lnx)(lnx^5)##
    = ##\ln x \frac{5x^4}{x^5} + \frac{\ln x^5 }{x}##
    = ln5 + ##lnx^4##
    = ##ln5x^4##
    ##\frac{dy}{dx}## = ##(x^5)^{lnx}## ##5lnx^4##
    Mod note: edited the line above to fix the exponent.
    I'm not confident in this answer but I've been trying all day and my brain hurts
     
    Last edited by a moderator: Aug 12, 2015
  12. Aug 12, 2015 #11

    RUber

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    You are good to here...but you should not include the 5x^4/x^5 in the logarithm.
    ##\ln x \frac{5x^4}{x^5} ## means ##\left( \frac{5x^4}{x^5} \right) \ln (x) = \frac{5\ln x}{x}##
    ##\frac{\ln x^5 }{x} =\frac{ 5\ln x}{x}##
    ##\ln (5x^4) \neq 5 \ln (x^4)##
     
  13. Aug 12, 2015 #12

    RUber

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    An alternate method would be to get the exponents out of the way early:
    lny=ln##(x^5)^{lnx}##
    lny=##lnxlnx^5 = 5 \ln x \ln x = 5 (\ln x) ^2 ##
    ##\frac{1}{y} \frac{dy}{dx}## = ##5 \frac{d}{dx}(\ln x)^2 ##
    Which may be a more straightforward application with the chain rule.
     
  14. Aug 12, 2015 #13
    I am more confused now? I did what you said earlier:

     
  15. Aug 12, 2015 #14
    lny=ln##(x^5)^{lnx}##
    lny=##lnxlnx^5 = 5 \ln x \ln x = 5 (\ln x) ^2 ##
    ##\frac{1}{y} \frac{dy}{dx}## = ##5 \frac{d}{dx}(\ln x)^2 ##
    ##\frac{dy}{dx}## = ##10(lnx).\frac{1}{x}##
    x=1
     
  16. Aug 12, 2015 #15
    Should it be ##\ln x . \frac{5x^4}{x^5} ##
    I am so confused
     
  17. Aug 12, 2015 #16

    RUber

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    This is right.
    You forgot your 1/y on the left side.
    I have no clue why you say x = 1.
    ##\frac1y \frac{dy}{dx}## = ##10(lnx).\frac{1}{x} =\frac{10\ln x }{x} ##
    ## \frac{dy}{dx} = y \frac{10\ln x }{x}##
    Then you are done.
     
  18. Aug 12, 2015 #17

    RUber

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    Yes. The chain rule says ##\frac{d}{dx} \ln x^5 = \frac{1}{x^5} \times 5x^4 ## Which is equal to ##\frac5x##
    This makes sense, since ##\frac{d}{dx} \ln x^5 = \frac{d}{dx} (5\ln x) = 5 \frac{d}{dx} \ln x = \frac5x##

    You had ##\ln x \times \frac{d}{dx} \ln x^5## which should be equal to ## \ln x \times \frac{1}{x^5} \times 5x^4##
     
  19. Aug 12, 2015 #18
    Thank you RUber. Sorry I am slow I need more practice. Really appreciate your help.
     
  20. Aug 12, 2015 #19

    RUber

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    Not a problem. Take your time and look for chances to simplify before jumping into the derivative. Practice makes us all better.
     
  21. Aug 16, 2015 #20
    Final answer attached. Thanks to all who helped.
     

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