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Implicit Differentiation

  1. Sep 28, 2009 #1
    Hi,

    So, I am reviewing Cal III, and I have come across something that I do not understand regarding implicit differentiation with partial derivatives:

    x^3 + y^3 + z^3 + 6xyz = 1

    implicit differentiation of z with respect to x:

    3x^2 + 3z^2(dz/dx) + 6yz + 6xy(dz/dx) = 0

    *notive the (dz/dx) are partial derivatives, not regular derivatives

    What I do not understand is that there are two '6xyz' derivatives. I understand how 6yz was formed, because it is wrt x, and the 6xy(dz/dx) because that is how the implicit part works, I believe. However, I do not understand how you can just add the variables twice, it seems like that would change this entire equality. . . ?

    Any help would be great!
    Thanks!
     
  2. jcsd
  3. Oct 2, 2009 #2

    lurflurf

    User Avatar
    Homework Helper

    That comes from the product rule.
    (6xyz)'=6(x)'yz+6xy(z')
    where x'=1, but z' stays around
     
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