# Composition and the chain rule

1. Sep 24, 2010

### Telemachus

1. The problem statement, all variables and given/known data
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.

2. Sep 24, 2010

### Susanne217

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

3. Sep 24, 2010

### Telemachus

Thank you Susanne

4. Sep 24, 2010

### Susanne217

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