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What are rules for $\alpha(\Delta x, \Delta y) , \beta(\Delta x, \Delta y)$ functions so that $f(x,y)=\mathrm{e}^{xy}$ to be differentiable on $\mathbb{R}^2$ by the following definition?

$\textbf{DEFINITION}$: Suppose that for every point $(a+\Delta x, b+\Delta y)$ in a deleted neighborhood of $(a,b)$ we can write: $f(a+\Delta x, b+\Delta y)-f(a,b)=A\Delta x+\Delta y +\alpha(\Delta x, \Delta y)\Delta x+\beta(\Delta x, \Delta y)\Delta y$ where $A,B$ are constant, and $\alpha(\Delta x, \Delta y) , \beta(\Delta x, \Delta y)$ are functions of $\Delta x, \Delta y$ such that $\displaystyle\lim_{(\Delta x, \Delta y)\to (0,0)}\alpha(\Delta x, \Delta y) =\lim_{(\Delta x, \Delta y)\to (0,0)}\beta(\Delta x, \Delta y) =0 $, then $f$ is said to be $\underline{differentiable}$ at $(a,b)$.

$\textbf{Remark}$: We should be find $A,B$ , and functions $\alpha(\Delta x, \Delta y) , \beta(\Delta x, \Delta y)$ at any point $(a, b)$ . Of course I know that $A=\dfrac{\partial f}{\partial x}(a,b) \ , \ B=\dfrac{\partial f}{\partial y}(a,b)$ . But What are $\alpha(\Delta x, \Delta y) , \beta(\Delta x, \Delta y)$ functions, exactly?

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I think you mean $A \Delta x + B \Delta y$? –  anonymous Dec 28 '12 at 14:40
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3 Answers

Let $A = be^{ab}$ and $B = ae^{ab}$. So we write $f(a + \Delta x, b + \Delta y) - f(a,b) - A \Delta x - B \Delta y = e^{(a+ \Delta x)(b + \Delta y)} - e^{ab} - \Delta x (be^{ab}) - \Delta y (ae^{ab})$.

By continuity, $e^{(a + \Delta x)(b + \Delta y)} \to e^{ab}$ as $\Delta x, \Delta y \to 0$. Also, $\Delta x(be^{ab}) - \Delta y(ae^{ab}) \to 0$ as $\Delta x, \Delta y \to 0$.

Since the right hand side tends to $0$, this establishes differentiability as per your definition.

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Try $$ f(x+\Delta x,y+\Delta y)-f(x+\Delta x,y)+f(x+\Delta x,y)-f(x,y) $$ and use the continuity of $e^{ax}$.

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You can expand the exponential as a Taylor series about a small quantity:

$$f(a+\Delta x,b+\Delta y) - f(a,b) = \exp{(a b)} [\exp{(b \Delta x + a \Delta y + \Delta x \Delta y)} - 1] $$

$$ = \exp{(a b)} \left [ b \Delta x + a \Delta y + \Delta x \Delta y + \frac{1}{2} (b \Delta x + a \Delta y )^2 + O(\Delta x,\Delta y)^3 \right ] $$

$$ =\exp{(a b)} \left [ b \Delta x + a \Delta y + \frac{1}{2} (b^2 \Delta x^2 + a^2 \Delta y^2 + 2 (1+a b) \Delta x \Delta y) + O(\Delta x,\Delta y)^3 \right ] $$

I believe this fits the definition you describe above and achieves the desired result in the limit as $\Delta x \rightarrow 0$ and $\Delta y \rightarrow 0$.

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