Composition and the chain rule

[Telemachus]

Homework Statement

I have a problem with the next exercise:

Given de function [tex]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}[/tex] with [tex]\vec{g}(t)=\begin{Bmatrix} x=at \\y=bt \end{matrix},t\in{\mathbb{R}}[/tex]

a) Find [tex]h=fog[/tex] y [tex]\displaystyle\frac{dh}{dt}[/tex] 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.

[tex]f_x(0,0)=0=f_y(0,0)[/tex]

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

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

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.

 

[Susanne217]

Homework Statement

I have a problem with the next exercise:

Given de function [tex]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}[/tex] with [tex]\vec{g}(t)=\begin{Bmatrix} x=at \\y=bt \end{matrix},t\in{\mathbb{R}}[/tex]

a) Find [tex]h=fog[/tex] y [tex]\displaystyle\frac{dh}{dt}[/tex] 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.

[tex]f_x(0,0)=0=f_y(0,0)[/tex]

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

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

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

 

[Telemachus]

Thank you Susanne

 

[Susanne217]

Thank you Susanne

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