1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partial derivative question

  1. Sep 13, 2007 #1

    duo

    User Avatar

    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

    User Avatar

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Partial derivative question
Loading...