# Thread: First and Second Partials at the Origin

1. ## First and Second Partials at the Origin

$f(x,y)=\sqrt[5]{x^5+y^5}$

The question is to find $f_x(0,0)$ and $f_{xx}(0,0)$

Solving $f_x(0,0)$ how I would normally take a partial derivative I get $\frac{x^4}{(x^5+y^5)^{\frac{4}{5}}}$ which is undefined at (0,0).

I get 1 when I use the limit definition of the partial derivative. I'm unsure how to calculate the the second partial though. I want to get an actual number and not 0/0 if possible.

2. Originally Posted by Shananay
$f(x,y)=\sqrt[5]{x^5+y^5}$

The question is to find $f_x(0,0)$ and $f_{xx}(0,0)$

Solving $f_x(0,0)$ how I would normally take a partial derivative I get $\frac{x^4}{(x^5+y^5)^{\frac{4}{5}}}$ which is undefined at (0,0).

I get 1 when I use the limit definition of the partial derivative. I'm unsure how to calculate the the second partial though. I want to get an actual number and not 0/0 if possible.
The formula $f_x(x,y) = \frac{x^4}{(x^5+y^5)^{\frac{4}{5}}}$ is valid whenever $x+y\ne0$. In particular, $f_x(x,0) = \frac{x^4}{(x^5)^{\frac{4}{5}}}=1$ when $x\ne0$. You have also shown (using the limit definition of the partial derivative) that $f_x(0,0) = 1$.

Therefore (again using the limit definition of the partial derivative) $f_{xx}(0,0) = \lim_{x\to0}\frac{f_x(x,0)-f_x(0,0)}x = \lim_{x\to0}\frac{1-1}x = \lim_{x\to0}\,\frac0x = 0$.

3. Originally Posted by Opalg
$\lim_{x\to0}\frac{f_x(x,0)-f_x(0,0)}x = \lim_{x\to0}\frac{1-1}x$
I'm not sure I understand this step. What exactly are the functions in the first numerator?

4. Originally Posted by Shananay
I'm not sure I understand this step. What exactly are the functions in the first numerator?
They are the values of the first x-derivative $f_x$ at the points (x,0) and (0,0). I'm using the fact that the second derivative $f_{xx}$ is the x-derivative of the first derivative.