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Some more logarithm stuff giving me trouble

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data

    y = 2^(sin x)

    Find derivative.

    2. Relevant equations



    3. The attempt at a solution

    [itex]y = 2^{sin x}[/itex]
    [itex]ln y = ln (2^{sin x})[/itex]
    [itex]ln y = sin x ln(2)[/itex]
    [itex]y' = [ln(2)] (cos x) (2^{sin x})[/itex]

    Yet, according to some rule in the book:

    a^x derivative is ln a times a^x, or in this case ln(2)2^(sin x)

    But usually these rules don't conflict with other valid ways of coming to the answer. Wolfram shows me my answer, mathway shows me the book's answer.
     
  2. jcsd
  3. Jul 24, 2011 #2

    Pengwuino

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    What you have is correct. Remember, you need to use the chain rule.

    [tex]{{d}\over{dx}}\left( a^{f(x)} \right) = ln(a)a^{f(x)}f'(x)[/tex]

    just as if it were something simpler like [itex]y = e^{x^2}[/itex]
     
  4. Jul 25, 2011 #3
    Thank you, I was getting frustrated. Another text book error, I suppose I shouldn't be surprised.
     
  5. Jul 25, 2011 #4

    Ray Vickson

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    I think you are mis-interpreting what the book said. According to you, the book said (d/dx) a^x = ln(a) * a^x, and that is perfectly true. However, your question was different. Are you saying that the book's answer for this specific problem, (d/dx) 2^(sin(x)) is wrong? Of course, books DO make mistakes, but I cannot figure out from what you said whether that is the case here, or not.

    RGV
     
  6. Jul 26, 2011 #5

    HallsofIvy

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    No, in this case, ln(2)2^(sin x) times the derivative of sin(x) so the derivative is ln(2)2^(sin(x))cos(x) exactly as you have.

     
  7. Jul 26, 2011 #6
    I'm sorry, I meant both actually.

    The book referenced the "box" where that rule was written (right above, on the same page) and then explicitly wrote the answer as n(2)2^(sin x) in accordance with that rule.

    I understand the rule is true, but it doesn't nullify the chain rule if I have some function for the exponent. That's what I gathered from the book's example, and that's what was the source of my frustration.

    As for mathway...

    [PLAIN]http://img263.imageshack.us/img263/1599/mathwaywrong.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  8. Jul 26, 2011 #7

    HallsofIvy

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    Then mathway is wrong. The derivative of [itex]2^{sin(x)}[/itex] is [itex]2^{sin(x)}ln(2)cos(x)[/itex]. The cos(x), the derivative of sin(x), is missing from what you show.
     
  9. Jul 26, 2011 #8
    Yes, this has been established.
     
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