Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $f(x,y)=\begin{cases}\frac{x^2y}{x^2+y^2} &\text{ if }(x,y) \neq (0,0)\\\\ 0&\text{ if }(x,y)=(0,0)\end{cases}$

and let $g(t)=(t,-2t).$ Prove that $f\circ g$ is differentiable in t=0, but the chain rule doesn't work.

share|cite|improve this question
Prove that $f \circ g$ is differentiable is easy. But I don't know how to prove that the chain rule doesn't work. – Henfe Jan 5 '13 at 19:52
Well, the chain rule would apply if $f$ and $g$ were differentiable. So you'd better show that at least one of them is not differentiable at the relevant point. – Chris Eagle Jan 5 '13 at 19:55
up vote 2 down vote accepted

I interpret the question like this: Since $f\circ g(t) = -\frac25t$, we have that $(f\circ g)'(t) = -\frac25$.

On the other hand, if the chain rule would be applicable, we would get

$$(f\circ g)'(0) = f'_x(0,0)\cdot 1 + f'_y(0,0) \cdot (-2) = 0,$$

since $f'_x(0,0) = f'_y(0,0) = 0$ as noted by vesszabo in his answer.

share|cite|improve this answer

$g$ is differentiable (it's easy, I hope :) ). So consider $f$. If it was diff'able at $0$, then we could write $$ f(x,y)-f(0,0)=A (x-0)+B(y-0)+R(x,y), $$ where $$ \lim_{(x,y)\to(0,0)} \frac{R(x,y)}{\sqrt{x^2+y^2}}=0. $$ A theorem says, in this case $A=D_1 f(0,0)$, $B=D_2 f(0,0)$, where $D_j$ denotes the partial derivative with respect to the $j$-th variable. Now using the definition of partial derivative we have $$ A=0,\quad B=0. $$ So we get $f(x,y)=R(x,y)$. Investigate the limit assumption. Introducing polar coordinates, $x=r\cos(\varphi)$, $y=r\sin(\varphi)$ we obtain $$ \lim_{(x,y)\to(0,0)} \frac{R(x,y)}{\sqrt{x^2+y^2}}=\lim_{r\to 0+} \frac{r^3\cos^2(\varphi)\sin(\varphi)}{r^3}=\cos^2(\varphi)\sin(\varphi), $$ which, in general, is not $0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.