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Is this derivative correct

  1. Aug 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Guys, is this right?
    And if it is, from the the y' got in there?

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

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  2. jcsd
  3. Aug 18, 2008 #2
    Re: Derivative

    You should use imageshack/other for images
  4. Aug 18, 2008 #3


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    Homework Helper

    Re: Derivative

    The answer looks reasonable - perhaps multiply the two negatives.
    I'm not sure what you mean by "And if it is, from the the y' got in there?"
  5. Aug 18, 2008 #4


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    Staff Emeritus
    Science Advisor

    Re: Derivative

    Here, z and y are both functions of some other variable, perhaps x or t. If z= f(y) and y is itself a function of x, then, by the chain rule
    [tex]\frac{dz}{dx}= \frac{dz}{dy}\frac{dy}{dx}[/tex]

    If, in particular, z= sin(1/y)= sin(y-1), and y is a function of x, then
    [tex]\frac{dz}{dx}= cos(1/y)(-1/y^2)\frac{dy}{dz}[/tex]
    or, the ' notation,
    z'= cos(1/y)(-1/y2)y'= (-cos(1/y)/y2)y'
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