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Partial derivatives and chain rule

  1. Oct 18, 2016 #1
    1. The problem statement, all variables and given/known data
    a. Given [itex]u=F(x,y,z)[/itex] and [itex]z=f(x,y)[/itex] find [itex]{ f }_{ xx }[/itex] in terms of the partial derivatives of of [itex]F[/itex].

    b. Given [itex]{ z }^{ 3 }+xyz=8[/itex] find [itex]{ f }_{ x }(0,1)\quad { f }_{ y }(0,1)\quad { f }_{ xx }(0,1)[/itex]

    2. Relevant equations
    Implicit function theorem, chain rule diagrams, Clairaut's theorem

    3. The attempt at a solution
    [itex]\frac { \partial z }{ \partial x } =-\frac { \frac { \partial u }{ \partial x } }{ \frac { \partial u }{ \partial z } } \quad \\ -\frac { { \partial }^{ 2 }z }{ { \partial x }^{ 2 } } =\frac { \partial }{ \partial x } \left[ \frac { \partial u }{ \partial x } { \left( \frac { \partial u }{ \partial z } \right) }^{ -1 } \right] \\ -\frac { { \partial }^{ 2 }z }{ { \partial x }^{ 2 } } =\left( \frac { { \partial }^{ 2 }u }{ \partial { x }^{ 2 } } +\frac { { \partial }^{ 2 }u }{ \partial z\partial x } +\frac { \partial z }{ \partial x } \right) { \left( \frac { \partial u }{ \partial z } \right) }^{ -1 }-{ \left( \frac { \partial u }{ \partial z } \right) }^{ -2 }\left( \frac { { \partial }^{ 2 }u }{ \partial x\partial z } +\frac { { \partial }^{ 2 }u }{ { \partial z }^{ 2 } } +\frac { \partial z }{ \partial x } \right) \left( \frac { \partial u }{ \partial x } \right) \\ \frac { { \partial }^{ 2 }z }{ { \partial x }^{ 2 } } ={ \left( \frac { \partial u }{ \partial z } \right) }^{ -2 }\left( { F }_{ xz }+\frac { { \partial }^{ 2 }u }{ { \partial z }^{ 2 } } \right) \left( \frac { \partial u }{ \partial x } \right) -\left( \frac { { \partial }^{ 2 }u }{ \partial { x }^{ 2 } } +{ F }_{ xz } \right) { \left( \frac { \partial u }{ \partial z } \right) }^{ -1 }[/itex] is this a correct attempt at part a? I'm not sure how to start part b
     
  2. jcsd
  3. Oct 18, 2016 #2

    PeroK

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    Why would the derivatives of ##f## depend on ##F##?

    Are you sure about this question?
     
  4. Oct 18, 2016 #3
    I'm not sure what you're asking; this is the problem statement. Are they related by the implicit function theorem?

    I think I made a mistake in lines 3 and 4 here's the correction:
    [itex]-\frac { { \partial }^{ 2 }z }{ { \partial x }^{ 2 } } =\left( \frac { { \partial }^{ 2 }u }{ \partial { x }^{ 2 } } +{ F }_{ xz }\frac { \partial z }{ \partial x } \right) { \left( \frac { \partial u }{ \partial z } \right) }^{ -1 }-{ \left( \frac { \partial u }{ \partial z } \right) }^{ -2 }\left( { F }_{ xz }+\frac { { \partial }^{ 2 }u }{ { \partial z }^{ 2 } } \frac { \partial z }{ \partial x } \right) \left( \frac { \partial u }{ \partial x } \right) \\ \frac { { \partial }^{ 2 }z }{ { \partial x }^{ 2 } } ={ \left( \frac { \partial u }{ \partial z } \right) }^{ -2 }\left( { F }_{ xz }-\frac { { \partial }^{ 2 }u }{ { \partial z }^{ 2 } } \frac { { F }_{ x } }{ { F }_{ z } } \right) \left( \frac { \partial u }{ \partial x } \right) -\left( \frac { { \partial }^{ 2 }u }{ \partial { x }^{ 2 } } -{ F }_{ xz }\frac { { F }_{ x } }{ { F }_{ z } } \right) { \left( \frac { \partial u }{ \partial z } \right) }^{ -1 }[/itex]
     
  5. Oct 18, 2016 #4

    PeroK

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    PS let ##f(x,y) = x^3 + y## and ##F(x, y, z) = 1##

    ##f_{xx} = 6x## yet all the derivatives of ##F## are zero.
     
  6. Oct 18, 2016 #5
    is this right?
    [itex]z[/itex] is a function of [itex]x[/itex] and [itex]y[/itex], so I can write [itex]F(x,y,f(x,y))=1[/itex], as long as [itex]x[/itex] and [itex]y[/itex] satisfies the relation.
    then I want the partial of [itex]z[/itex] with respect to [itex]x[/itex]
    [itex]F(x,y,f(x,y))=1\\ \frac { \partial }{ \partial x } F=0\\ \frac { \partial F }{ \partial x } +\frac { \partial F }{ \partial z } \frac { \partial z }{ \partial x } =0\\ \frac { \partial z }{ \partial x } =-\frac { \frac { \partial F }{ \partial x } }{ \frac { \partial F }{ \partial z } } [/itex] now f is in terms of F
     
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