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Implicit Differenciation

  1. Oct 26, 2009 #1
    i have a relation z4 + x2z = x5 ,
    now i differenciate it implictly to find dz/dx ,
    but if we observe carefully we have a relation between x and z , but not a function of x ,
    so is this derivative correct or even if implicit Differenciation valid over here ???????
     
  2. jcsd
  3. Oct 26, 2009 #2

    arildno

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    An equation can IMPLICITY define a function!
    This is what is contained in the inplicit function theorem.

    Suppose you have a two-variable function:
    [tex]G(x,z)=z^{4}+x^{2}z-x^{5}[/tex]
    Note that if we look at the equation:
    [tex]G(x,z)=0(*)[/tex]
    we see that the solutions of this equation are the same as the ones to your original equation.

    Furthermore, the implicit function theorem states that, under suitable conditions, there exists a function Z(x), so that for ALL x, (*) is satisfied:
    [tex]G(x,Z(x))=0[/tex]

    Since this holds for all x, we may differentiate our expression (with respect to x), getting:
    [tex]\frac{\partial{G}}{\partial{x}}+\frac{\partial{G}}{\partial{z}}\frac{dZ}{dx}=0[/tex]

    From this equation, you should be able to solve for dZ/dx.
     
  4. Oct 26, 2009 #3

    HallsofIvy

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    In the case that z is NOT a function of x, your example, or, simpler, [itex]x^2+ z^2= 1[/itex], solving for z would give two or more expressions give "z as a function of x". Then dz/dx, at a specific value of x, would depend upon which of the values of z for that x you chose. Since differentiating both sides of the formula with respect to x, [itex]4z^3 z'+ 2xz+ x^2z'- 5x^4= 0[/itex] for your case or [itex]2x+ 2zz'= 0[/itex] for [itex]x^2+ z^2= 1[/itex] still have both x and z, in order to solve for z' you would have to put in not only the value of x, but the correct value for z. In otherwords, you have to choose one of the possible values and so are reducing to a "single valued" function.
     
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