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Implicit differentiation and the product rule

  1. Jun 7, 2010 #1
    An example of implicit differentation in Stewart, 6th ed, p 883, is given as follows:

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

    Differentiating to find dz/dx,

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

    where the product rule was used to differentiate 6xyz with respect to x.

    Why isn't the product rule also used when differentiating explicitly?


    if z = xy, dz/dx = y.

    rather than dz/dx = dx/dx(y) + dy/dx(x).

  2. jcsd
  3. Jun 7, 2010 #2


    Staff: Mentor

    In the differentiation above, z is assumed to be an implicit function of x but y is not.
    As above, z is assumed to be an implicit function of x, but y is not. If z = xy, then by the product rule, dz/dx = x * dy/dx + d(x)/dx * y.

    Since y is NOT assumed to be a function of x, dy/dx = 0, so dz/dx = x * 0 + 1 * y = y.

    Another way to approach this, not using the product rule, is by realizing that if y is not a function of x, it is essentially a constant as far as differentiation with respect to x is concerned. So d(xy)/dx = y * d(x)/dx = y. (I'm using the idea that d(kf(x))/dx = k * f'(x).)
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