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Partial Derivatives

  1. Nov 8, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-11-8_20-24-17.png

    So I know I have to take the derivative with respect to x, then respect to y, then respect to z, but I am not getting the right answer. I know that the answer is 0 and my professor did it with very few steps that I do not understand. Can someone please guide me through it?
     
  2. jcsd
  3. Nov 8, 2016 #2
    The trick is to recognize that the partial derivative is a linear operator, that is, the partial derivative of a sum is equal to the sum of the partial derivatives of the individual terms:
    [tex]\frac{\partial}{\partial x} \left( f + g \right) = \frac{\partial f}{\partial x} + \frac{\partial g}{\partial x} [/tex]
    Try splitting ##f(x,y,z)## into two parts, take the partial derivatives separately and see where that leads you.
     
  4. Nov 8, 2016 #3
    So if I take the partial derivative wrt x and the result has no x term what does that mean?
     
  5. Nov 8, 2016 #4
    Could you provide an explicit example of what you're trying to ask?
     
  6. Nov 8, 2016 #5
    So I split the function like you said and I got xln(y100+37z11 / xz rad(y2+1 and when I take the partial derivative of that term wrt x, x is no longer in that term
     
  7. Nov 8, 2016 #6
    The partial derivative means to take all other variables as constants - what happens when you differentiate a constant?
     
  8. Nov 8, 2016 #7
    I did take the other variables as constant. I'm saying that by keeping everything constant but x, the x term differentiates to 1 if you look at the function.
     
  9. Nov 8, 2016 #8
    How does it "differentiate to ##1##"? Take a closer look at it:
    [tex]\frac{x \ln(y^{100} + 37z^{11})}{x z \sqrt{y^2 +1}} = \frac{\ln(y^{100} + 37z^{11})}{z \sqrt{y^2 +1}} [/tex]
    is independent of ##x##. So what happens when you take the partial derivative wrt ##x##?
     
  10. Nov 8, 2016 #9
    Isn't that what you would get if you differentiate with respect to x?
     
  11. Nov 8, 2016 #10
    No, I haven't differentiated - take a closer look: all I did was to divide both the numerator and denominator by ##x##.
     
  12. Nov 8, 2016 #11
    So that differentiates to 0 then
     
  13. Nov 8, 2016 #12
    Yup. Now look at the second term and see if you notice something similar.
     
  14. Nov 8, 2016 #13
    So for the second term, differentiating to z is 0 as well
     
  15. Nov 8, 2016 #14
    Yup. So immediately we conclude that ##f_{xyz}## is indeed ##0##.
     
  16. Nov 8, 2016 #15
    But we haven't taken the partial derivative wrt y. Don't you have to go in order taking the partial derivative of x, y, then z?
     
  17. Nov 8, 2016 #16
    For a suitably well-behaved function, it doesn't matter - as the hint to the question itself explicitly says. Even if you wanted to "go in order", notice that if I had a function that depends only on ##x## and ##y## but not ##z##, when I perform the partial derivatives wrt ##x## and ##y##, I will still end up with a result that is independent of ##z## i.e. I won't have a ##z## suddenly appearing in my expression. So in the end, when I take the partial derivative wrt ##z##, it will still be ##0##.
     
  18. Nov 8, 2016 #17
    I think my confusion lies in splitting the function into two parts. So for the first part of the function differentiating wrt x gives 0 but that's not the case for the second part of the function. Why do I ignore that second part for x?
     
  19. Nov 8, 2016 #18
    Like, wouldn't I differentiate the entire function in terms of x?
     
  20. Nov 8, 2016 #19
    What we are actually doing by splitting is to evaluate separately and then add the result together.
    Suppose we split a function ##f(x,y,z)## into two parts, say
    [tex]f(x,y,z) = g(x,y,z) + h(x,y,z)[/tex]
    Then, because partial derivation is linear, we have
    [tex]f_{xyz} = g_{xyz} + h_{xyz}[/tex]
    so we can find ##g_{xyz}## and ##h_{xyz}## separately and combine the results to get ##f_{xyz}##.
     
  21. Nov 8, 2016 #20
    Ohh ok I got it now. Thank you so much for bearing with me.
     
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