# Different Definition of Differentiability

Question from Calculus 3:

1. Let $f(x,y)$ be a function that's defined in a neighborhood of $(x_0,y_0)$. Show that exist $a(x,y), b(x,y)$ that are continuous at $(x_0, y_0)$, such that:

$f(x,y)=f(x_0, y_0)+a(x,y)(x-x_0)+b(x,y)(y-y_0)$.

I was able to solve this, but am writing this to explain the next part of the question which I couldn't solve.

1. Do there exist any points $(x_1, y_1)\neq (x_0 ,y_0)$, such that $b(x_1,y_1)=f_y(x_1,y_1)$. This I wasn't able to solve, though I found that $b(x,y)=f_y (x_0,y_0)$.

For more background, the way I calculated $b$, was through the definition of differentiability, when $f(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+r(x,y)$, when $r(x,y)=o(d((x,y), (x_0,y_0))$, and using a theorem, I wrote:

$r(x,y)=u(x,y)(x-x_0)+v(x,y)(y-y_0)$, when $lim_{(x,y)\rightarrow(x_0,y_0)}v(x,y)=0$, and I defined $b(x,y)=f_y(x_0,y_0)+v(x,y)$.

So to say that $b(x_1,y_1)=f_y(x_1,y_1)$ is to say that $f_y(x_1,y_1)=f_y(x_0,y_0)+v(x_1,y_1)$. Now does there exist $(x_1,y_1)\neq (x_0,y_0)$, such that this is true?

Consider e.g. $f(x,y) = y^2$ with $x_0 = 0$, $y_0 = 0$. Then you can take $a(x,y) = 0$ and $b(x,y) = y$. You never have $y_1 = f_y(x_1,y_21) = 2 y_1$ unless $y_1 = 0$.
EDIT: OK, here's a counterexample. Take \eqalign{f(x,y) &= x^2 y + {y}^{3}\cr a(x,y) &= x y\cr b(x,y) &= y^2 \cr x_0 = y_0 = 0\cr} Then $f_y(x_1,y_1) = x^2 + 3 y^2 \ne b(x,y) = y^2$ unless $x=y=0$.
• This definitely isn't a counter example since for all $\varepsilon >0$, we can take $|x|<\varepsilon$, such that $||(x,y)||<\varepsilon$, and therefore, in every neighborhood of $(0,0)$, exists $(x,y)$ in that neighborhood such that the theorem is true. Are you trying to say that the theorem is true? – NSA Mar 28 '16 at 14:33