- #1

Sho Kano

- 372

- 3

## Homework Statement

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]

## Homework Equations

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

## 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