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What to use and how to best do it?

  1. Aug 31, 2014 #1
    Given: cos(x)[itex]\frac{1}{x}[/itex]

    Find [itex]\frac{dy}{dx}[/itex]:

    Now I know... [itex]\frac{dy}{dx}[/itex] of cos(x) = -sin(x)

    and [itex]\frac{dy}{dx}[/itex] of [itex]\frac{1}{x}[/itex]= [itex]\frac{-1}{x^2}[/itex]

    But I am unsure what formula to apply, and where to go from here?
     
  2. jcsd
  3. Aug 31, 2014 #2

    PeroK

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    Try using a logarithm.
     
  4. Aug 31, 2014 #3
    I get confused when I convert it to logarithmic.
    E.g I get the base as cos(x) and inside the log as y =1/x
    So...
     
  5. Aug 31, 2014 #4

    PeroK

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    [tex]y = cos(x)^{\frac{1}{x}}[/tex]
    [tex]ln(y) = ln(cos(x)^{\frac{1}{x}})= \frac{1}{x}ln(cos(x))[/tex]
    Now differentiate both sides using the chain rule.
     
  6. Aug 31, 2014 #5
    Thanks very much for your help

    Much appreciated !
     
  7. Aug 31, 2014 #6

    Mark44

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    Since y doesn't appear in your first line above, it doesn't make much sense to talk about dy/dx. If we know that y = (cos(x))1/x, then dy/dx is meaningful.
    I understand what you're trying to say, but you're not saying it correctly. You don't take "dy/dx" of something. dy/dx is already the derivative of y (with respect to x). To indicate that you want to take the derivative with respect to x of cos(x), write d/dx(cos(x)).
    You mean d/dx(1/x).
     
  8. Aug 31, 2014 #7

    && the quotient rule. Correct?
     
  9. Aug 31, 2014 #8

    PeroK

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    I never use the quotient rule.
     
  10. Aug 31, 2014 #9
    Than what of ##x^{-1} * ln(cos(x))##??
    That's a product. Shouldn't either product or quotient rule be applied?
     
  11. Aug 31, 2014 #10

    PeroK

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    I always use the product rule and chain rule. The quotient rule is too much to remember. And, unnecessary.
     
  12. Aug 31, 2014 #11
    I do as well. I just stated it as the quotient rule because of the way you wrote it.
     
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