# Differentiation question

1. Dec 15, 2008

### boombaby

1. The problem statement, all variables and given/known data
Define f(0,0)=0 and $$f(x,y)=\frac{x^{3}}{x^{2}+y^{2}}$$ if (x,y)!=(0,0)
Let $$\gamma$$ be a differentiable mapping of R1 into R2, with $$\gamma(0)=(0,0)\;and\; |\gamma'(0)>0|$$. Put $$g(t)=f(\gamma(t))$$ and prove that g is differentiable for every t in R1

2. Relevant equations

3. The attempt at a solution
I'm not having a strong understanding of differentiation of vector functions yet, so I'm not really sure if my proof is valid, check it please. Thanks!
f(x,y) is not differentiable at (0,0) so chain rule fails. If $$lim_{t->0} \frac{g(t)}{t} exists$$(since g(0)=0), it's done.
$$\frac{g(t)}{t}=\frac{f(\gamma(t))}{t}=\frac{\gamma^{3}_{1}(t)}{t(\gamma^{2}_{1}(t)+\gamma^{2}_{2}(t))}$$. when t approaches zero, the denominator is not zero since $$|\gamma'(0)|>0$$, the numerator can be applied the mean value theorem and becomes $$\gamma^{3}_{1}(t)-\gamma^{3}_{1}(0)=3\gamma^{2}_{1}(0)\gamma'_{1}(0)t=0\;since\;\gamma_{1}(0)=0$$. So g'(0) exists and equals to 0.

Thanks a lot!

actually...I'm a little more confident with it now

Last edited: Dec 15, 2008
2. Dec 15, 2008

### boombaby

okay now. Is this proof right?

Last edited: Dec 16, 2008