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Homework Help: Implicit Differentiation (for a normal derivative) with Multivariable Functions

  1. Nov 11, 2011 #1
    1. The problem statement, all variables and given/known data


    Define z as a function of x, find z'.

    2. Relevant equations

    I guess the two equations above...

    3. The attempt at a solution

    Well, I just differentiated the first one with respect to x and got:

    z' = [itex]\frac{-1-z^{3}{3z[itex]^{2}[/itex]x}[/itex]

    If I differentiate the second equation, I get something completely different...

    z' = [itex]\frac{2x-z}{x}[/itex]

    I'm sorry if I typed some of my work above wrong, but the main reason I'm stuck on this question is: I don't even see why I need the two equations to solve this problem

    Thanks in advance to you guys.
    Last edited: Nov 11, 2011
  2. jcsd
  3. Nov 11, 2011 #2

    I like Serena

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    Homework Helper

    Welcome to PF, paul2211! :smile:

    Your variable y is a function of x and z.
    As such you need to consider it as well when differentiating.

    Note that in this case you can use your second equation to eliminate y altogether.

    Btw, you appear to have made at least one type when differentiating the first equation.
  4. Nov 11, 2011 #3
    Thank you for your quick reply Serena.

    So does this mean y = y(x, z(x))?

    If so, and I somehow couldn't cancel the y between the two equations, wouldn't I have [itex]\frac{\partial y}{\partial x}[/itex], [itex]\frac{\partial y}{\partial z}[/itex] and [itex]\frac{dz}{dx}[/itex] to solve for?

    Since I would only have two equations, does this mean I can't solve for [itex]\frac{dz}{dx}[/itex] if I didn't have the insight to cancel out the other variable? (y in this case).
  5. Nov 11, 2011 #4

    I like Serena

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    Yes and yes.

    If you didn't cancel y beforehand, y will probably cancel itself as long as you apply the proper derivative rules.
    It's just less work to cancel y beforehand.
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