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Differentiation of logarithms

  1. Jun 15, 2015 #1
    • Member warned about not using the template and showing work only in an image
    the variables x and y are positive and related by x^a.y^b=(x+y)^(a+b) where a and b are positive constants. By taking logarithms of both sides, show that dy/dx=y/x. provided that bx not equal to ay.
     

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  3. Jun 15, 2015 #2

    ehild

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    You should have typed in your derivation. In that case, I would be able to show where your error is. The differentiation is not correct.
     
  4. Jun 15, 2015 #3
    x^a.y^b=(x+y)^(a+b)
    => ln(x^a.y^b) = ln((x+y)^(a+b))
    =>a lnx + b lny = (a+b)ln(x+y)
    =>a. d/dx lnx + b. d/dx lny = (a+b). d/dx ln(x+y)
    => a/x + b/y. dy/dx = (a+b)/(x+y). dy/dx

    The last line is where I think I made a mistake. Can you please help?
     
  5. Jun 15, 2015 #4

    ehild

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    Yes, d/dx [ln(x+y)]=1/(x+y) d/dx(x+y) . What is d/dx(x+y)? it is a sum, you have to differentiate both terms. What is dx/dx?
     
  6. Jun 15, 2015 #5

    Ray Vickson

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    I know the question tells you to use logarithms, but there is a much easier way: the constraint ##x^a y^b = (x+y)^{a+b}## implies that
    [tex] \left(\frac{x}{x+y}\right)^a \left( \frac{y}{x+y} \right)^b = 1, [/tex]
    hence ##r^a (1-r)^b = 1##, where ##r = x/(x+y)##. For given ##a,b##, that means that ##r## is the solution, or one of two solutions to an equation, so ##x/(x+y)## is a constant. Thus, ##y = cx## for some constant ##c>0##.
     
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