# Partial derivatives and chain rule

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1. Oct 18, 2016

### Sho Kano

1. The problem statement, all variables and given/known data
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 ${ f }_{ x }(0,1)\quad { f }_{ y }(0,1)\quad { f }_{ xx }(0,1)$

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

3. 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

2. Oct 18, 2016

### PeroK

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

3. Oct 18, 2016

### Sho Kano

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:
$-\frac { { \partial }^{ 2 }z }{ { \partial x }^{ 2 } } =\left( \frac { { \partial }^{ 2 }u }{ \partial { x }^{ 2 } } +{ F }_{ xz }\frac { \partial z }{ \partial x } \right) { \left( \frac { \partial u }{ \partial z } \right) }^{ -1 }-{ \left( \frac { \partial u }{ \partial z } \right) }^{ -2 }\left( { F }_{ xz }+\frac { { \partial }^{ 2 }u }{ { \partial z }^{ 2 } } \frac { \partial z }{ \partial x } \right) \left( \frac { \partial u }{ \partial x } \right) \\ \frac { { \partial }^{ 2 }z }{ { \partial x }^{ 2 } } ={ \left( \frac { \partial u }{ \partial z } \right) }^{ -2 }\left( { F }_{ xz }-\frac { { \partial }^{ 2 }u }{ { \partial z }^{ 2 } } \frac { { F }_{ x } }{ { F }_{ z } } \right) \left( \frac { \partial u }{ \partial x } \right) -\left( \frac { { \partial }^{ 2 }u }{ \partial { x }^{ 2 } } -{ F }_{ xz }\frac { { F }_{ x } }{ { F }_{ z } } \right) { \left( \frac { \partial u }{ \partial z } \right) }^{ -1 }$

4. Oct 18, 2016

### 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.

5. Oct 18, 2016

### Sho Kano

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