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Homework Help: Partial Derivatives

  1. Oct 13, 2008 #1
    Find [tex]\frac{\partial y}{\partial x}[/tex] of [tex]3sin (x^2 + y^2) = 5cos(x^2 - y^2)[/tex]

    [tex]\partial y = 6ycos(x^2 + y^2) - 10ysin(x^2 - y^2)[/tex]
    [tex]\partial x = 6xcos(x^2 + y^2) + 10xsin(x^2 - y^2)[/tex]

    so I thought [tex]\frac{\partial y}{\partial x} = \frac{6ycos(x^2 + y^2) - 10ysin(x^2 - y^2)}{6xcos(x^2 + y^2) + 10xsin(x^2 - y^2)}[/tex]

    but instead, the answer is supposed to be:

    [tex]\frac{\partial y}{\partial x} = - \frac{6xcos(x^2 + y^2) + 10xsin(x^2 - y^2)}{6ycos(x^2 + y^2) - 10ysin(x^2 - y^2)}[/tex]
  2. jcsd
  3. Oct 13, 2008 #2


    Staff: Mentor

    I don't see how you went from an equation to two partial differentials. What happened to the '=' in your first equation?

    Instead, take the partial with respect to x of both sides, keeping in mind that the partial of x with respect to x is 1. You'll have an equation that involves x, y, and partial of y w.r.t. x, and you should be able to solve for this partial by using algebra.

  4. Oct 13, 2008 #3
    isnt [tex]3sin (x^2 + y^2) = 5cos(x^2 - y^2) \Rightarrow 3sin (x^2 + y^2) - 5cos(x^2 - y^2) = 0[/tex]
  5. Oct 13, 2008 #4


    Staff: Mentor

    Yes, the first equation implies the second. In your post, you skipped a lot of steps between where you started and the next line.

    If you take the partial w.r.t x of 3 sin(x^2 + y^2) - 5 cos(x^2 - y^2), you're going to have factors of partial wrt x of (x^2 + y^2) and of (x^2 - y^2). For the first, you should get
    2x + 2y*partial y w.r.t x. (Sorry I'm not able to use Tex, the database seems very slow and cranky today.) The other factor is similar, but with - between the two terms.
    To summarize, after you take partials w.r.t. x of both sides of the equation, you should get this:
    3 cos(x^2 + y^2)*(2x + 2y*partial y w.r.t. x) + 5 sin(x^2 - y^2) * (2x + 2y*partial y w.r.t. x) = 0.

    You should be able to solve algebraically for partial y w.r.t. x.

    An important point is that if you start with an equation, and apply an operator to both sides, you end up with an equation.
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