Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

General differential question

  1. Oct 2, 2014 #1
    Hey guys. I have a question regarding the differential operator d.

    Say we have an equation d(Z+X*Y^2)=0
    If we want to differentiate the expression in the parenthesis, are these the steps to follow?
    d(Z+X*Y^2)=0
    dZ+d(X*Y^2)=0

    Apply product rule to the second term:
    dZ+Y^2*dX+X*dY^2=0

    Here is where I get confused. To simply the 3rd term (X*dY^2), is the simplification this:
    2Y*X*dY or 2*X*dY?
     
  2. jcsd
  3. Oct 2, 2014 #2

    Mark44

    Staff: Mentor

    Sort of. Assuming that f is a function of x, y, and z, then the total differential df involves the three partials.
    In other words, $$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz$$
    You should get this:
    $$\frac{\partial (xy^2)}{\partial x} dx + \frac{\partial (xy^2)}{\partial y} dy $$
    For each partial, treat the other variable as if it were a constant. Is that clear?
     
  4. Oct 2, 2014 #3
    Got it. Thank you.

    I don't know how to write out in the format you did, but for that last term partial(x*y^2)/partialy*dy,
    I was a little confused on how that simplifies.
    We hold x constant for that term, so this gives:
    x*partial(y^2)/partial_y*dy
    Does this give
    x*2*y*partial(y)/partial_y*dy
    or
    x*2*partial(y)/partial_y*dy?
     
  5. Oct 2, 2014 #4

    Mark44

    Staff: Mentor

    This -- x*2*y*partial(y)/partial_y*dy -- which simplifies to 2xy dy. The partial of y with respect to y is just 1.

    I wrote my previous reply using LaTeX, which isn't too difficult. It looks like this:
    \frac{\partial f}{\partial x}
    Put a pair of $ symbols at front and back, and it renders like this:
    $$\frac{\partial f}{\partial x}$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: General differential question
Loading...