# Understanding Partial derivative proof

Here is the Proposition and the beginning of the proof:

$$\begin{array}{l}{\text { Proposition 9.2 Let } X \text { be an open set in } \mathbb{R}^{m} \text { that is a product of open }} \\ {\text { intervals, let } f: X \rightarrow \mathbb{R} \text { be a real-valued function on } X, \text { and let } M \text { be a }} \\ {\text { positive constant. Suppose that }} \\ {\qquad\left|\frac{\partial f}{\partial x_{i}}\right| \leq M}\end{array}$$

$$\begin{array}{l}{\text { throughout the open set } X \text { for } i=1,2, \ldots, m . \text { Then }} {|f(\mathbf{u})-f(\mathbf{v})| \leq \sqrt{m} M|\mathbf{u}-\mathbf{v}|} \\{\text { for all points } \mathbf{u} \text { and } \mathbf{v} \text { of } X}\end{array}$$

$$\begin{array}{l}{\text { Proof Let } \mathbf{u}=\left(u_{1}, u_{2}, \ldots, u_{m}\right) \text { and } \mathbf{v}=\left(v_{1}, v_{2}, \ldots, v_{m}\right) . \text { The fact that } X \text { is }} \\ {\text { a product of open intervals guarantees that there exist real numbers } a_{i} \text { and } b_{i}} \\ {\text { for } i=1,2, \ldots, m \text { such that } a_{i}

$$\begin{array}{l}{\text { where }} \\ {\qquad w_{k, i}=\left\{\begin{array}{ll}{v_{j}} & {\text { if } i \leq k} \\ {u_{j}} & {\text { if } i>k}\end{array}\right.} \\ {\text { Then } a_{i}

$$\frac{d}{d t}\left(f\left((1-t) \mathbf{w}_{k-1}+t \mathbf{w}_{k}\right)\right)=\left.\left(v_{k}-u_{k}\right) \frac{\partial f}{\partial x_{k}}\right|_{(1-t) \mathbf{w}_{k-1}+t \mathbf{w}_{k}}$$

$$(v_k-u_k)$$ occurs as a consequence of the chain rule. But what does the $$x_k$$ and the bar notation of the following segment represent?

$$\left.\frac{\partial f}{\partial x_{k}}\right|_{(1-t) \mathbf{w}_{k-1}+t \mathbf{w}_{k}}$$

It means you find the partial derivative of $$f$$ with respect to $$x_k$$ (this is now a function of $$\mathbf{x}$$) and you plug in $$\mathbf{x}=(1-t)\mathbf{w}_{k-1}+t\mathbf{w}_k$$.