**Proof plan.** We decompose the increment $\delta f = f(x + \delta x, y + \delta y) - f(x, y)$ by moving along coordinate axes one at a time, apply single-variable Taylor expansions along each axis, and take the [limit](/page/Limit) as $\delta x, \delta y \to 0$ to identify the differential. **Step 1: Decompose the increment along axes.** Write the increment as a sum of two segments: \begin{align*} \delta f &= f(x + \delta x, y + \delta y) - f(x, y) \\ &= \bigl[f(x + \delta x, y + \delta y) - f(x + \delta x, y)\bigr] + \bigl[f(x + \delta x, y) - f(x, y)\bigr]. \end{align*} The first bracket varies $y$ at fixed $x + \delta x$; the second varies $x$ at fixed $y$. **Step 2: Apply single-variable Taylor expansions.** Treating each bracket as a [function](/page/Function) of one variable and applying [Taylor's Theorem](/theorems/827) to first order: \begin{align*} f(x + \delta x, y) - f(x, y) &= \delta x \frac{\partial f}{\partial x}(x, y) + o(\delta x), \\ f(x + \delta x, y + \delta y) - f(x + \delta x, y) &= \delta y \frac{\partial f}{\partial y}(x + \delta x, y) + o(\delta y). \end{align*} Expanding $\frac{\partial f}{\partial y}(x + \delta x, y)$ around $x$ using Taylor again: \begin{align*} \frac{\partial f}{\partial y}(x + \delta x, y) = \frac{\partial f}{\partial y}(x, y) + o(1). \end{align*} **Step 3: Combine and identify the differential.** Substituting: \begin{align*} \delta f = \delta x \frac{\partial f}{\partial x}(x, y) + \delta y \frac{\partial f}{\partial y}(x, y) + o(\delta x) + o(\delta y) + \delta x \delta y \cdot \frac{\partial^2 f}{\partial x \partial y}(x, y) + o(\delta x \delta y). \end{align*} As $\delta x, \delta y \to 0$, the higher-order terms vanish, giving the differential: \begin{align*} df = \lim_{\delta x, \delta y \to 0} \delta f = dx \frac{\partial f}{\partial x} + dy \frac{\partial f}{\partial y}. \end{align*}