# How to prove this theorem about differentiability of a multivariable function?

The theorem says that:

A function $f: \mathbb{R}^2 \to \mathbb{R}$ is differentiable at $(x_0, y_0)$ if its partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are continuous at $(x_0, y_0)$.

How do I prove this? I know that for a function to be differentiable, the condition is:

$$\displaystyle \lim_{\lVert h,k \rVert \to 0} \dfrac{f(x_0+h, y_0+k) - f(x_0, y_0) - h\,\frac{\partial f}{\partial x}\rvert_{(x_0, y_0)} - k\,\frac{\partial f}{\partial y}\rvert_{(x_0, y_0)}}{\lVert h,k \rVert} = \lim_{\lVert h,k \rVert \to 0} \mathcal O(h,k) = 0$$

The problem is that the definition of differentiability is using the values of partial derivatives at the point itself, and not the function. I am not understanding how I can "link" that statement to the definition of continuity of partial derivatives, which are functions themselves. I am not even getting where I could start!

• Tip: This obviously is (to me at least) a standard result. Have you tried looking in a book or googling? Sep 10 '17 at 10:31

Let us write \begin{align*} f(x_0 + h, y_0 + k) - f(x_0,y_0) &= f(x_0 + h,y_0) - f(x_0,y_0)\\ &\qquad \quad {}+ f(x_0 + h, y_0 + k) - f(x_0 + h,y_0). \end{align*} Now, we can apply the Mean Value Theorem to the two pairs of terms: \begin{align*} f(x_0 + h, y_0) - f(x_0,y_0) &= h \cdot \frac{\partial f}{\partial x}(b_1,y_0)\\ f(x_0 + h, y_0 + k) - f(x_0 + h,y_0) &= k \cdot \frac{\partial f}{\partial y} (x_0+h,b_2) \end{align*} for some $b_1 \in (x_0,x_0 + h)$ and $b_2 \in (y_0, y_0 + k)$. Therefore, \begin{align*} & \frac{\left| f(x_0+h,y_0+k)-f(x_0,y_0)-h\frac{\partial f}{\partial x}(x_0,y_0)-k\frac{\partial f}{\partial y}(x_0,y_0) \right|}{\| (h,k)\|}\\ ={} & \frac{\left| h\left( \frac{\partial f}{\partial x}(b_1,y_0)-\frac{\partial f}{\partial x}(x_0,y_0) \right) - k \left( \frac{\partial f}{\partial y}(x_0+h,b_2) - \frac{\partial f}{\partial y}(x_0,y_0) \right) \right|}{\| (h,k)\|}\\ \leq{} & \frac{| h |}{\| (h,k) \|} \cdot \left| \frac{\partial f}{\partial x}(b_1,y_0)-\frac{\partial f}{\partial x}(x_0,y_0) \right| + \frac{|k |}{\| (h,k) \|} \cdot \left| \frac{\partial f}{\partial y}(x_0+h,b_2) - \frac{\partial f}{\partial y}(x_0,y_0) \right| \\ \leq{} & \left| \frac{\partial f}{\partial x}(b_1,y_0)-\frac{\partial f}{\partial x}(x_0,y_0) \right| + \left| \frac{\partial f}{\partial y}(x_0+h,b_2) - \frac{\partial f}{\partial y}(x_0,y_0) \right| \end{align*} since $| h | / \| (h,k) \|$ and $| k | / \| (h,k) \|$ are less than or equal to $1$ for all $(h,k) \neq (0,0)$.
Now, taking $\lim_{\|(h,k)\| \to 0}$, we get $0$, because $(b_1,y_0) \to (x_0,y_0)$ and $(x_0+h,b_2) \to (x_0,y_0)$ as $(h,k) \to (0,0)$, and the partial derivatives are continuous at $(x_0,y_0)$. This is where we use the continuity of the partial derivatives at $(x_0,y_0)$.