# Partial derivatives and chain rule

• Sho Kano
In summary: I can use the chain rule to find the partial derivatives of f with respect to x and yIn summary, using the implicit function theorem and chain rule, we can find the second partial derivative of z in terms of the partial derivatives of F. Additionally, by setting up an implicit function such that F(x,y,f(x,y))=1, we can use the chain rule to find the partial derivatives of f with respect to x and y.
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

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

Are you sure about this question?

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

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.

PeroK said:
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

## 1. What are partial derivatives?

Partial derivatives are derivatives of a multivariable function with respect to one of its variables, holding all other variables constant. They represent the rate of change of the function with respect to that specific variable.

## 2. What is the chain rule in the context of partial derivatives?

The chain rule is a calculus rule that allows us to find the derivative of a composite function. In the context of partial derivatives, it is used to find the partial derivative of a multivariable function with respect to a specific variable, by taking into account the chain of dependencies between the variables.

## 3. How is the chain rule applied to find partial derivatives?

To apply the chain rule in finding partial derivatives, we first take the derivative of the outer function with respect to the variable of interest. Then, we multiply it by the derivative of the inner function with respect to the same variable. This process is repeated for each variable in the function.

## 4. What is the difference between a partial derivative and an ordinary derivative?

A partial derivative measures the rate of change of a multivariable function with respect to one of its variables, while holding all other variables constant. An ordinary derivative, on the other hand, measures the rate of change of a single-variable function with respect to its input.

## 5. In what fields of science are partial derivatives and the chain rule commonly used?

Partial derivatives and the chain rule are commonly used in fields such as physics, engineering, economics, and finance. They are also essential in areas of math such as calculus, differential equations, and vector calculus.

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