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

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).)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Implicit differentiation product Date
B Implicit Differentiation Jun 12, 2017
B Implicit differentiation or just explicit? Mar 26, 2017
I Implicit differentiation Dec 12, 2016
I Implicit differentiation Sep 29, 2016
B (ASK) Implicit Differentiation Aug 17, 2016