How does one show that the function is differentiable?

[Feynman's fan]

I'd like to show that if $\alpha>\frac{1}{2}$ then $(x^2+y^2)^\alpha$ is differentiable at $(0,0)$.

The usual way is to show that the partial derivatives are continuous at $(0,0)$.

Yet I am a little confused how to show that $2x\alpha(x^2+y^2)^{\alpha-1}$ is continuous at $(0,0)$. I have tried working it out by definition, yet it seems to be a mess.

Any hints are very appreciated!

 

[scurty]

I'd like to show that if $\alpha>\frac{1}{2}$ then $(x^2+y^2)^\alpha$ is differentiable at $(0,0)$.

The usual way is to show that the partial derivatives are continuous at $(0,0)$.

Yet I am a little confused how to show that $2x\alpha(x^2+y^2)^{\alpha-1}$ is continuous at $(0,0)$. I have tried working it out by definition, yet it seems to be a mess.

Any hints are very appreciated!

I might be wrong, so you should wait until others give advice, but I believe something that can get you started would be to first break this into two cases for the values of $\alpha$. After that, I would use a squeeze theorem argument for one of the cases.

 

[Simon Bridge]

I'd be inclined to use the limit definition of the partial derivative.
http://mathinsight.org/partial_derivative_limit_definition

 

[Feynman's fan]

Simon Bridge,
thank you, it's all clear now!

 

[Simon Bridge]

No worries - happy New Year.