Multivariable calculus - explain what the teacher did

Question

The teacher gave this exercise:

Find $D_f(a)$ when $f: \mathbb R^n \to \mathbb R$, $f(x)=<x,\xi>^2$ where $\xi \in \mathbb R^n$.

What I did:

I wrote it as $$f(x)= (\sum_{i=1}^{n}x_i \xi_i)^2$$

and so $$\frac{df}{dx_j} = 2\xi_j(\sum_{i=1}^{n}x_i \xi_i)$$

What the teacher did (please explain, I don't understand what he did):

"we will find $D_f(a)$ using how it acts in direction $h$":

$$f(x)= (\sum_{i=1}^{n}x_i \xi_i)^2$$

So:

$$D_f(a)h=\frac{d}{dt}|_{t=0}f(a+th) = ...=2(\sum_{i=1}^n a_i \xi_i)(\sum_{i=1}^nh_i \xi_i)$$

And so it follows that:

$D_f(a)h=<a,\xi><h,\xi>$

QED

Could someone explain what he did? as this is not even similar to my answer

Answer 1

You computed the gradient $\nabla f(x)=\left({\partial f(x)\over\partial x_1},\ldots,{\partial f(x)\over\partial x_n}\right)$ and obtained $$\nabla f(x)=2\langle x,\xi\rangle\>\xi\ .$$ Now the "official" $df(x)$ is a linear functional, and the gradient $\nabla f(x)$ is the vector that represents this functional via the scalar product. This means that $$df(x).h=\bigl\langle\nabla f(x), h\bigr\rangle=2\langle x,\xi\rangle\>\langle\xi,h\rangle\ .$$ Therefore the teacher obtained the same as you did, but he came to an end, whereas your answer is somehow "dangling".

Answer 2

Managed to solve it and understand what the teacher did:

$$D_f(a)h = \lim_{t \to 0} \frac{f(a+th)-f(a)}{t} =\lim_{t \to 0} \frac {<a+th,\xi>^2-<a,\xi>^2}{t} = \lim_{t \to 0} \frac{(<a,\xi>+<th,\xi>)(<a,\xi>+<th,\xi>)-<a,\xi>^2}{t}=\lim_{t \to 0} \frac{<a,\xi>^2+2t<a,\xi><h,\xi>+t^2<h,\xi> - <a,\xi>^2}{t} =\lim_{t \to 0} \frac{2t<a,\xi><h,\xi>+t^2<h,\xi>}{t} = 2<a,\xi><h,\xi> $$