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Natural Log in a derivative

  1. Mar 10, 2012 #1

    MacLaddy

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    1. The problem statement, all variables and given/known data

    I have a problem that I'm working on that I have almost solved, yet I am just a tad off of what the book says the answer is. I will show the way I'm doing it, and where I depart from the steps the book takes.

    The graph of [itex]y=(x^2)^x[/itex] has two horizontal tangent lines. Find equations for both of them. (I bolded "two" because this is part of my mistake)


    2. Relevant equations

    Chain rule, exponential rule, etc

    3. The attempt at a solution

    [itex]y=(x^2)^x = x^{2x}[/itex]

    [itex]\ln{y}=\ln{x^{2x}}[/itex]

    This is where the book and I deviate. This is what I do.

    [itex]lny = 2xlnx[/itex]

    Taking the derivative

    [itex]\frac{1}{y}y'=2x\frac{1}{x}+2lnx[/itex]

    [itex]\frac{1}{y}y'=2+2lnx[/itex]

    [itex]y'=y(2+2lnx)[/itex]

    [itex]y'=x^{2x}(2+2lnx)[/itex]

    Solving for zero

    [itex]x^{2x} = 0[/itex] when[itex] x=0[/itex]

    [itex]2+2lnx = 0[/itex]

    [itex]2lnx = -2[/itex]

    [itex]lnx = \frac{-2}{2}[/itex]

    [itex]lnx = -1[/itex]

    [itex]e^{lnx} = e^{-1}[/itex]

    [itex]x = e^{-1} = 1/e[/itex]

    Here is where the problem lies, the book never brought the exponent in front, like [itex]lny = 2xlnx[/itex]. Instead it kept [itex]lny = lnx^{2x}[/itex] and then found the derivative, which ends up with two answers.

    What are the rules here? Any advice (about this problem) is appreciated.

    Mac
     
    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2

    Ray Vickson

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    The function (x^2)^x is well-defined for x < 0 or x > 0, but the function exp(2*x*ln(x)) is not defined for x < 0: you would be taking the log of a negative number. The two expressions are the same only if x > 0.

    RGV
     
    Last edited: Mar 10, 2012
  4. Mar 10, 2012 #3

    MacLaddy

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    Thanks for the fast reply, Ray, but I am not sure I understand.

    Are you saying that because [itex]x^{2x}[/itex] is not defined as greater than zero, then you can not change [itex]lnx^{2x}[/itex] to [itex]2xlnx[/itex]?

    If that's true, then why does that exponent rule apply wrt Ln? I think I'm still missing something.
     
    Last edited: Mar 10, 2012
  5. Mar 10, 2012 #4

    SammyS

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    First of all, [itex](x^2)^x = x^{2x}[/itex] only for x ≥ 0.

    More generally you can write, [itex](x^2)^x = |x|^{2x}\,,[/itex] which is true for all real values of x.

    Therefore, [itex]\displaystyle \ln(y)=\left\{ \begin{array}{rcl}
    2x\ln(x)& \text{ if } & x\ge 0 \\
    \\
    2x\ln(-x)& \text{ if } & x<0\end{array}\right.[/itex]

    Work with that.
     
  6. Mar 10, 2012 #5

    MacLaddy

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    Sorry guys, but this isn't making any sense to me. Is there a reference you can point me to?

    I understand that for ln(x), x > 0, but I'm not seeing how that applies to the exponent, and why in your piecewise function above you can put a negative in the Natural Log. I also do not understand why x for [itex]x^{2x}[/itex] must be equal to or greater than zero.

    The rule as I've always seen it is [itex]\ln{a^b} = b*lna[/itex]
     
    Last edited: Mar 10, 2012
  7. Mar 10, 2012 #6

    SammyS

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    Right.

    If x = -3/2, then [itex]\displaystyle (x^2)^x = \left(\left(\frac{-3}{2}\right)^2\right)^{-3/2}=\left(\frac{9}{4}\right)^{-3/2}=\frac{8}{27}\,,[/itex] whereas, [itex]\displaystyle x^{2x} = \left(\frac{-3}{2}\right)^{2(-3/2)}=\left(\frac{-3}{2}\right)^{-3}=\frac{-8}{27}\,,[/itex]
     
  8. Mar 10, 2012 #7

    HallsofIvy

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    Then you need to go back an review. What you should have seen was [itex]\ln{a^b}= b\ln{a}[/itex] if a> 0[/itex]. You keep ignoring the possibility that a is negative, in which case you cannot take the derivative.
     
  9. Mar 10, 2012 #8

    SammyS

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    If x itself is a negative number, as in, x = -5, then ln(-x) = ln(-(-5)) = ln(5).

    That's how you can have ln(-x) .
     
  10. Mar 10, 2012 #9

    MacLaddy

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    Ah, I see. I was not aware of this rule.



    Yes, definitely need to review. I just want to make sure I know exactly what to review.

    That makes more sense, thanks for the clarification.

    I am going to dig into this question for a bit, and hopefully I will understand it. I think I have some serious deficiencies in my understanding somewhere. I'm just not sure if it goes back to algebraic rules I should know, or if this is something new in calculus that I'm not grasping.

    I appreciate both of your help. If I do have more questions, would you recommend starting a new thread, or continuing with this one?
     
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