Composition and the chain rule

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Homework Help Overview

The discussion revolves around the differentiability of a function defined as f(x,y) with specific conditions at the point (0,0) and the application of the chain rule in this context. The original poster is attempting to find the derivative of a composition of functions involving f and a parameterized path defined by g(t).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster explores the differentiability of f at (0,0) and questions whether the chain rule can be applied given that f is not differentiable at that point. They discuss the existence of partial derivatives and the limit behavior of f as (x,y) approaches (0,0).

Discussion Status

Participants are engaged in examining the conditions under which the chain rule can be applied, particularly focusing on the differentiability of the function involved. The original poster has raised pertinent questions about the implications of differentiability on the composition of functions.

Contextual Notes

The original poster references a course resource related to the chain rule for functions of two variables, indicating an exploration of external materials to aid understanding.

Telemachus
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Homework Statement


I have a problem with the next exercise:

Given de function f(x,y)=\begin{Bmatrix} \displaystyle\frac{xy^2}{x^2+y^2} & \mbox{ if }& (x,y)\neq{(0,0)}\\0 & \mbox{if}& (x,y)=(0,0)\end{matrix} with \vec{g}(t)=\begin{Bmatrix} x=at \\y=bt \end{matrix},t\in{\mathbb{R}}

a) Find h=fog y \displaystyle\frac{dh}{dt} for t=0

The thing is that I've found that f isn't differentiable at (0,0). The partial derivatives exists at that point, I've found them by definition.

f_x(0,0)=0=f_y(0,0)

And then I saw if it was differentiable at that point.

\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{xy^2}{(x^2+y^2)^{3/2}}}

In the polar form it gives that this limit doesn't exists, so it isn't differentiable at that point. So I can't apply the chain rule there, right?

To ensure the differentiability of a composed function, both function must be differentiable. If one isn't, then the composition isn't differentiable at a certain point. Right?

Bye there, and thanks.
 
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Telemachus said:

Homework Statement


I have a problem with the next exercise:

Given de function f(x,y)=\begin{Bmatrix} \displaystyle\frac{xy^2}{x^2+y^2} & \mbox{ if }& (x,y)\neq{(0,0)}\\0 & \mbox{if}& (x,y)=(0,0)\end{matrix} with \vec{g}(t)=\begin{Bmatrix} x=at \\y=bt \end{matrix},t\in{\mathbb{R}}

a) Find h=fog y \displaystyle\frac{dh}{dt} for t=0

The thing is that I've found that f isn't differentiable at (0,0). The partial derivatives exists at that point, I've found them by definition.

f_x(0,0)=0=f_y(0,0)

And then I saw if it was differentiable at that point.

\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{xy^2}{(x^2+y^2)^{3/2}}}

In the polar form it gives that this limit doesn't exists, so it isn't differentiable at that point. So I can't apply the chain rule there, right?

To ensure the differentiability of a composed function, both function must be differentiable. If one isn't, then the composition isn't differentiable at a certain point. Right?

Bye there, and thanks.

Found this here course where they deal with the chainrule on two variable function...
http://www.math.oregonstate.edu/hom...lculusQuestStudyGuides/vcalc/chain/chain.html
 
Thank you Susanne
 
Telemachus said:
Thank you Susanne

You are welcome. I find it better to see worked through examples something to understand the methods and theory :)
 

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