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Derivative of compound ln()

  1. Jun 26, 2015 #1
    1. The problem statement, all variables and given/known data
    $$y=a\cdot x\cdot ln\left(\frac{b}{x}\right)$$
    The derivative should be 0 (to maximize), what's x?
    2. Relevant equations
    $$(ln\:x)'=\frac{1}{x}$$
    $$(x^a)'=ax^{(a-1)}$$
    $$(uv)'=u'v+v'u$$

    3. The attempt at a solution
    $$\dot y=a \left[ ln \left( \frac{b}{x} \right)-x\frac{x}{b}x^{-2} \right]=a \left[ ln \left( \frac{b}{x} \right)-\frac{1}{b} \right]$$
    $$\dot y=0 \rightarrow ln \left( \frac{b}{x} \right)-\frac{1}{b}=0 \rightarrow x=e^{\left( \frac{1}{b} \right)}$$
    But the answer should be ##x=\frac{b}{e}##
     
    Last edited: Jun 26, 2015
  2. jcsd
  3. Jun 26, 2015 #2

    SteamKing

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    Have you tried rewriting ln (b / x) using the rules of logarithms before taking the derivative of the whole expression?
     
  4. Jun 26, 2015 #3

    SammyS

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    You missed the Chain rule.

    Also it may be make things easier to first expand the logarithm.

    ##\displaystyle \ \ln\left(\frac{b}{x}\right)=\ln(b)-\ln(x) \ ##
     
  5. Jun 26, 2015 #4
    This question is from my post in physics, i re-wrote the constants in order not to confuse.
    Yes, i used there the logarithm's rules:
    $$y=a\cdot x\cdot ln\left(\frac{b}{x}\right)=a\cdot x\cdot (ln\:b-ln\:x)$$
    $$\dot y=a\left[( ln\:b-ln\:x )-x\frac{1}{x} \right]=a(ln\:b-ln\:-1)=a\left[ ln\left( \frac{b}{x}\right)-1 \right]$$
     
  6. Jun 26, 2015 #5
    $$\left[x\cdot ln\left( \frac{b}{x} \right)\right]'=ln \left( \frac{b}{x} \right)+x\frac{x}{b}(-b)\frac{1}{x^2}=x\left[ ln\left( \frac{b}{x} \right)-1\right]$$
    I think it's solved, i forgot the constant b, thanks
     
    Last edited: Jun 26, 2015
  7. Jun 26, 2015 #6

    SammyS

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    What's the derivative of ##\displaystyle \ \frac{b}{x} \ ?\ ##
     
    Last edited: Jun 26, 2015
  8. Jun 26, 2015 #7
    See my answer in #5, i solved, thanks
     
  9. Jun 26, 2015 #8

    SammyS

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    That's incorrect.
     
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