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Partial Derv.

  1. Oct 23, 2003 #1
    All right, I know that partial derv. are not all that hard. All you do in lets say f(x)= x^2+2yx+3y is find the dervitives of x while you treat y as a constent, and vice versa. But I keep running into problems having square roots. I hat these things.
    One example could be f(x)= sqr(20-x^2-7y^2). I know yo treat the y as a constent, but I am horrible with my derv. Any help would be appreciated.
  2. jcsd
  3. Oct 23, 2003 #2
    If you hate square roots so much, would you be more comfortable with implicit differentiation? Can you easily take the deriviatve of g(x)=20-x2-7y2? Sure you can. Now, g(x)=[f(x)]2 and 2ff'=g' so f'=g'/(2f) which agrees with what you know about square roots. You know f, so all you need is g', which is easy to find.
    So the partial derivative of f wrt x is,

    Please forgive the sloppy notation.
  4. Oct 23, 2003 #3


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    The other way is to apply the chain rule to the power rule:

    (d/dx) (u^n) = n u^(n-1) (du/dx)

    then just plug in the right things for u and n and finish the rest of the work
  5. Oct 23, 2003 #4
    Hello, Arden1528!

    You said, "I keep running into problems having square roots.
    One example could be f(x,y)= sqr(20 - x^2 - 7y^2)"

    Just where is your difficulty?
    You know how to take partial derivatives.
    You know that a square root is a one-half power.
    And you know the Chain Rule.

    f(x,y) = (20 - x2 - 7y2)1/2

    Hence, fx = (1/2)(20 - x2 - 7y2)-1/2(-2x) = -x/(20 - x2 - 7y2)1/2

    And, fy = {1/2)(20 - x2 - 7y2)-1/2(-14y) = -7y/(20 - x2 - 7y2)1/2
  6. Oct 24, 2003 #5
    I did not even think about just putting the problems to the (1/2) power. I seem to forget the things that would make these problems easier. But thanks for all the help. I feel like I just forgot a really elementry operation...thanks
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