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Homework Help: Partial derivative question

  1. Sep 13, 2007 #1

    duo

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    1. The problem statement, all variables and given/known data

    d/dx 1/sin(y/2)


    3. The attempt at a solution

    this isn't an entire question, just looking for clarification about something.

    i have been asked as part of a larger question to find the partial derivative of 1/sin(y) with respect to x. in this case you treat y as a constant, yes?

    so d/dx of (sin(y/2) = cos(y/2)*dy/dx, since y is treated as a constant, dy/dx = 0, so d/dx of sin(y/2) = 0. if you use the quotient rule on the full equation, you divide by 0^2, which is 0, so you end up dividing by 0. Is this correct?

    The full expression actually has

    sin(x/2 + y/2)

    above instead of 1. is it just a matter of rearranging this expression so that the bottom cancels out? or does d/dx sin(y/2) not actually evaluate to 0 in the first place?

    thanks in advance!
     
  2. jcsd
  3. Sep 13, 2007 #2

    I think you are making this way too complicated as it shouldn't be.

    Well first of all, you mentioned about dividing by 0^2. Remember that you can't divide anything by 0.

    If you were asked to take the partial derivative of

    [tex]
    \sin(\frac{x}{2}+\frac{y}{2})
    [/tex]

    with respect to x, as you mentioned earlier, just treat y as a constant in this case, then take derivative.

    For instance, derivative for [tex]\sin(\frac{x}{2}+\frac{1}{3})[/tex] is simply [tex]\frac{1}{2}cos(\frac{x}{2}+\frac{1}{3})[/tex]
     
  4. Sep 13, 2007 #3

    duo

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    right, but it's [tex]\frac{\sin({\frac{x}{2} + \frac{y}{2}})}{\sin({y/2})}[/tex]

    with respect to x. so you use the quotient rule to evaluate it.

    but the expression on the bottom evaluates to 0 when you differentiate with respect to x? or am i totally wrong? also thank you for responding ;-)
     
  5. Sep 13, 2007 #4
    You don't necessarily have to take the derivative of bottom because it's a constant. Kind of like integral. Pull the constant out.

    For instance, can you find the derivative of

    [tex]\frac{\sin({\frac{x}{2} + \frac{1}{4}})}{\sin({1/4})}[/tex]

    It is the exact same idea, except y is your constant here.
     
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