Differential of scalar product

Question

Task from homework:

Let $f:\Bbb R^n\to\Bbb R$, $f(x,y)=\langle x, y\rangle$, where $\langle\cdot,\cdot\rangle$ means the scalar product in $\Bbb R^n$. Find the differential $Df(x,y)(h,k)$.

First, the domain of $f$ is surely wrong, so with correcting it to $\Bbb R^n\times \Bbb R^n$, I'm struggling to even start because every theorem we ever mentioned in class was about functions with the domain in $\Bbb R^n$. How can $f$ be partially differentiated if the components $x$ and $y$ are again vectors? Partial derivatives were my first idea, but any help would be appreciated.

Answer 1

The value of the differential at a point $(x,y)$ is the linear part of the difference $\;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity: $$f(x+h, y+k)-f(x,y)=\langle x+h,y+k\rangle -\langle x, y\rangle=\underbrace{\langle x,k\rangle+\langle h, y\rangle}_\text{linear terms} +\underbrace{\langle h, k\rangle}_{o(\|(h,k)\|)},$$ so $\; Df_{(x,y)}(h,k)=\langle x,k\rangle+\langle h, y\rangle$.

Answer 2

The (usual, euclidean) scalar product is defined via $$ \langle x ,y \rangle = x_1 y_1 + x_2 y_2 + \dots + x_n y_n $$ with $x,y \in \mathbb{R}^n$, so your function is essentially $ f:\mathbb{R}^n \times \mathbb{R}^n \simeq \mathbb{R}^{2n} \to \mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + \dots + x_n y_n. $$From here you should be able to conclude.