Partial derivatives and chain rule

[Sho Kano]

Homework Statement

a. Given u=F(x,y,z) and z=f(x,y) find { f }_{ xx } in terms of the partial derivatives of of F.

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

Homework Equations

Implicit function theorem, chain rule diagrams, Clairaut's theorem

The Attempt at a Solution

\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 } is this a correct attempt at part a? I'm not sure how to start part b

 

[PeroK]

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

Are you sure about this question?

 

[Sho Kano]

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

Are you sure about this question?

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:
-\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 }

 

[PeroK]

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.

 

[Sho Kano]

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.

is this right?
z is a function of x and y, so I can write F(x,y,f(x,y))=1, as long as x and y satisfies the relation.
then I want the partial of z with respect to x
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 } } now f is in terms of F