I have been trying to solve the following question without success.
Question:
Show that $f:\mathbb{R^n} \times \mathbb{R^n}\to \mathbb{R}$, defined as $$f(x,y) = x\cdot y$$ is differentiable and find its derivative at any $(x,y) \in \mathbb{R^n} \times \mathbb{R^n}$.
What I tried to do:
Heavily inspired by Differentiability of matrix-vector product map, I'm trying to understand how that would work here. I'm pretty sure that this general result could be used here, but I'm still trying to understand two things:
1) How one would get, from the definition, that the denominator ought to be $\lVert (h,H) \rVert$
2) How does one get $\|B(h,k)\|\leq\|B\|\|h\|\|k\|=o(\|(h,k)\|).$ It seems to me that one would use the following result: If $A\in L(R^n,R^m)$ and $A\in L(R^m,R^k)$, then
$$\lVert BA \lVert\leq \lVert B\lVert\lVert A\lVert$$ but I'm unsure how that would go here.
Recall that a function $f : U \to \mathbb{R}$ (where $U$ is an open subset of $\mathbb{R}^m$) is said to be differentiable at $x \in U$ if there exists a linear transformation $T: \mathbb{R^m} \to \mathbb{R}$ such that $\lim_{h\to 0}\frac{f(x+h) - f(x) - Th}{\|h\|} = 0$, in which case we call $T$ the derivative of $f$ at $x$.
In our case, the derivative at $(x_0, y_0) \in \mathbb{R}^n \times \mathbb{R}^n$ will be a linear transformation $T : \mathbb{R^n}\times\mathbb{R}^n \to \mathbb{R}$.
For $(h,k) \in \mathbb{R}^n \times \mathbb{R}^n$, define, $T(h, k) = y\cdot h + x\cdot h$. Now check the limit
$$\lim_{(h,k)\to \mathbb{0}}\frac{f(x+h,y+k) - f(x,y) - Th}{\|(h, k)\|} = \lim_{(h,k)\to \mathbb{0}}\frac{(x + h)\cdot(y+k) - x\cdot y - (y \cdot h + x\cdot k)}{\|(h,k)\|} = \lim_{(h,k)\to \mathbb{0}}\frac{x\cdot y + x\cdot k + h \cdot y + h\cdot k - x\cdot y - (y \cdot h + x\cdot k)}{\|(h, k)\|} = \lim_{(h,k)\to \mathbb{0}}\frac{h\cdot k}{\|(h, k)\|}$$
Now, $|h\cdot k| \leq \|h\|\|k\| \leq \begin{cases} \|h\|^2 \quad \text{if} \quad \|h\| \geq \|k\| \\ \|k\|^2 \quad \text{if} \quad \|h\| \leq \|k\| \end{cases} $
where the first inequality is Cauchy-Schwartz. So, $|h\cdot k| \leq \|h\|^2 + \|k\|^2$. It follows that $$0 \leq \frac{|h\cdot k|}{\|(h, k)\|} \leq \frac{\|h\|^2 + \|k\|^2}{\|(h, k)\|} = \|(h, k)\|$$
So, by Sandwich theorem, we get, $\lim_{(h,k)\to \mathbb{0}}\frac{h\cdot k}{\|(h, k)\|} = 0$.
(This proof was a generalization of the proof given in Spivak's Calculus on Manifolds for differentiability of product function $p: \mathbb{R^2} \to \mathbb R$ defined as $p(x, y) = xy$.)