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Derivative of a logarithm

  1. Dec 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello, I have to calculate the derivative of [itex]y = log_x (x+1)[/itex] so I used the formula of the derivative of a n-base logarithm and I get [itex]y' = 1/((x+1)logx)[/itex] but that's wrong, why ?

    Thanks

    2. Relevant equations
    [itex]log_a x = 1/(xlog(a))[/itex]
     
  2. jcsd
  3. Dec 3, 2013 #2

    dextercioby

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    Use that log_a {b} = ln{b}/ln{a}.
     
  4. Dec 3, 2013 #3
    But why that formula doesn't work ?
     
  5. Dec 3, 2013 #4

    dextercioby

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    Well, just raise your formula to the power a and use that a^{something} = e^{ln a (something)}
     
  6. Dec 3, 2013 #5

    HallsofIvy

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    That's wrong because you have assumed a formula, [tex]d(log_a(x))/dx= 1/(x ln(a))[/tex], that is true for constant base, is also true when the base is a variable. The fact that it is a variable means that base adds its own "variation" to the derivative.

    I would start with [itex]y= log_x(x+ 1)[/itex], then write [itex]x+ 1= x^y[/itex]. Now differentiate both sides with respect to x: [itex]1= x^{-y}ln(x)(dy/dx)+ x[/itex] so that [itex]dy/dx= (1- x)x^y/ln(x)[/itex]
     
  7. Dec 3, 2013 #6

    jedishrfu

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    so now you would have y=ln(x+1)/ln(x) and so you'd use the product rule on y = ln(x+1) * (1/ln(x)).
     
  8. Dec 3, 2013 #7
    So why the base changing formula is valid for a variable base too ?
     
  9. Dec 4, 2013 #8

    jedishrfu

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    It's valid for any base logarithm the fact that in your example it's a variable base doesn't matter. What's nice is that it take the variable base out of the picture and makes your problem somewhat simpler.
     
  10. Dec 5, 2013 #9

    HallsofIvy

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    Why wouldn't it be? Each value of a variable is a number so what ever is true for a number is true for each value of the variable. It is only when you are doing things that involve the way the variable changes, such as taking a derivative or an integral, that you have to take into account that it is an integral.
     
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