# Equivalence of definitions of multivariable differentiability

The usual definition of differentiability for a multivariable function says that $f:\mathbb{R}^2 \to \mathbb{R}$ is differentiable at $(x,y)$ if there is a linear map $d_{(x,y)} f : \mathbb{R}^2\to \mathbb{R}$ such that

$$\lim_{(h,k)\to 0} \frac{f(x+h,y+k) - f(x,y) - d_{(x,y)}f(h,k)}{\Vert(h,k)\Vert} = 0.$$

This is equivalent to asking that

$$f(x+h,y+k) = f(x,y) + d_{(x,y)}f(h,k) + \epsilon(h,k) \Vert(h,k)\Vert$$

where $\epsilon(h,k)\to 0$ as $(h,k)\to 0$.

However, the calculus book that I'm teaching out of this semester defines differentiability differently, asking that

$$f(x+h,y+k) = f(x,y) + d_{(x,y)}f(h,k) + \epsilon_1(h,k) h + \epsilon_2(h,k) k$$

where $\epsilon_1(h,k)\to 0$ and $\epsilon_2(h,k)\to 0$ as $(h,k)\to 0$. Why is this an equivalent definition? I see why the second implies the first: the obvious definition $\epsilon = \frac{\epsilon_1 h + \epsilon_2 k}{\Vert(h,k)\Vert}$ works because $\frac{|h|}{\Vert(h,k)\Vert} \le 1$ and $\frac{|k|}{\Vert(h,k)\Vert} \le 1$. But I don't see the converse: given $\epsilon(h,k)$, how do I define $\epsilon_1$ and $\epsilon_2$?

• Try $\Vert(h,k)\Vert=\Vert(h,k)\Vert^2/\Vert(h,k)\Vert$. Commented Apr 10, 2014 at 14:27
• @TonyPiccolo - Thanks! If you wrote that out in more detail and submitted it as an answer, I'd accept it. Commented Apr 11, 2014 at 13:43
• Thank you but I ask you only the title of the calculus book. Commented Apr 11, 2014 at 18:56

Try $\Vert(h,k)\Vert=\Vert(h,k)\Vert^2/\Vert(h,k)\Vert$. – Tony Piccolo Apr 10 at 14:27