# Differentiability and the chain rule (multivariable calculus)

• Sep 24th 2010, 10:33 AM
Ulysses
Differentiability and the chain rule (multivariable calculus)
I have a problem with the next exercise:

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

a) Find $\displaystyle h=fog$ y $\displaystyle \displaystyle\frac{dh}{dt}$ 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.

$\displaystyle f_x(0,0)=0=f_y(0,0)$

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

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

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.
• Sep 24th 2010, 12:50 PM
Opalg
I think this problem is easier than you are making it. In fact, $\displaystyle h(t) = \dfrac{(at)(bt)^2}{(at)^2+(bt)^2} = \dfrac{abt}{a^2+b^2}$ (and this formula holds also when t=0, because $\displaystyle h(0) = f(0,0) = 0$). Therefore $\displaystyle h'(t) = \frac{ab}{a^2+b^2}$ for all t, including t=0, whether or not f is differentiable at (0,0).
• Sep 24th 2010, 02:52 PM
Ulysses
Yes, you're right. Unless for that part of the problem. But I forgot to tell that then it asks me to use the chain rule too. I've made this part this way, and arrived to a similar conclusion, I think you've made a little mistake (you have forgotten the square for b), lets see:
$\displaystyle h(t)=\begin{Bmatrix} \displaystyle\frac{ab^2t^3}{a^2t^2+b^2t^2} & \mbox{ if }& (t)\neq{0}\\0 & \mbox{if}& t=0\end{matrix}$

Then using the limit definition we got $\displaystyle \dysplaystyle\frac{dh(0)}{dt}=\dysplaystyle\frac{a b^2}{a^2+b^2}$

But using the chain rule it gives 0, so the chain rule evidently doesn't work on this case, and thats because the function "f" isn't differentiable at the given point.

Thanks.
• Sep 25th 2010, 12:17 AM
Opalg
Quote:

Originally Posted by Ulysses
Yes, you're right. Unless for that part of the problem. But I forgot to tell that then it asks me to use the chain rule too. I've made this part this way, and arrived to a similar conclusion, I think you've made a little mistake (you have forgotten the square for b), lets see:
$\displaystyle h(t)=\begin{Bmatrix} \displaystyle\frac{ab^2t^3}{a^2t^2+b^2t^2} & \mbox{ if }& (t)\neq{0}\\0 & \mbox{if}& t=0\end{matrix}$

Then using the limit definition we got $\displaystyle \dysplaystyle\frac{dh(0)}{dt}=\dysplaystyle\frac{a b^2}{a^2+b^2}$

But using the chain rule it gives 0, so the chain rule evidently doesn't work on this case, and thats because the function "f" isn't differentiable at the given point.

Thanks.

You're right, of course, I should have written $\displaystyle \frac{ab^2}{a^2+b^2}$. The partial derivatives of f at (0,0) are both zero, so the chain rule gives the wrong answer for h'(0), and the reason for that is that f is not differentiable at (0,0) so the chain rule does not apply.