I have question about presentation of directional derivative.
Sometimes I see directional derivates and gradient to be represented as a function but I think that it is convenient to use "vector presentation" instead.

My question is that is there some cases and/or good reasons why should use functional presentation instead of vector?
To make my point clear I wrote an example what I mean about "presentations".
Simply I mean the presentation of vector, using comma $\displaystyle (2,1)$ versus using i and j components $\displaystyle 2i + j$

There is one example twice, for "function presentation" and for "vector presentation".

Let we have a function: $\displaystyle f(x,y) = x^2 - y^2 $ and we calculate gradient and directional derivative of function $\displaystyle f$.


Function presentation:

Gradient (I include all phases here):
$\displaystyle \nabla f(x,y) = \frac{\partial f}{\partial x} i+ \frac{\partial f}{\partial y}j $

$\displaystyle = \frac{\partial}{\partial x} (x^2 - y^2)i+ \frac{\partial}{\partial y}(x^2 - y^2)j $

$\displaystyle = (2x+0)i - (0-2y)j$

$\displaystyle = \underline{\underline{2xi - 2yj}}$

Next we calculate directional derivative of function f at this point (2,-1) to the direction of $\displaystyle \overline{\imath} - 3\overline{\jmath}$

The vector $\displaystyle \overline{\imath} - 3\overline{\jmath}$ has magnitude $\displaystyle \sqrt{1^2 + (-3)^2} = \sqrt{10}$. The unit vector in the direction $\displaystyle \overline{\imath} - 3\overline{\jmath}$ is thus $\displaystyle \hat{u} = \frac{1}{\sqrt{10}} (\overline{\imath} - 3\overline{\jmath})$

Directional derivative is then
$\displaystyle D_{\hat{u}} f(a,b) = \hat{u} \cdot \nabla f = \frac{1}{\sqrt{10}} (i -3j) \cdot (2xi-2yj)$

$\displaystyle =\frac{1}{\sqrt{10}} [1 \cdot 2x + (-3) \cdot (-2y)] = \frac{1}{\sqrt{10}} (2x + 6y)$

and plugging the point (2,-1) we get
$\displaystyle =\frac{1}{\sqrt{10}} (4-6) = \underline{\underline{\frac{-2}{\sqrt{10}}}}$


Vector presentation


$\displaystyle \nabla f(x,y) = \frac{\partial f}{\partial x} i , \frac{\partial f}{\partial y}j$

$\displaystyle = \frac{\partial}{\partial x} (x^2 - y^2)i , \frac{\partial}{\partial y}(x^2 - y^2)j $

$\displaystyle = (2x,2y)$

Next we calculate directional derivative of function f at this point (2,-1) to the direction of $\displaystyle (1,- 3)$

Let's plugin the point to the Gradient: $\displaystyle \underline{\underline{\nabla f(2,-1) = (4,2)}}$

The vector $\displaystyle (1,- 3)$ has magnitude $\displaystyle \sqrt{1^2 + (-3)^2} = \sqrt{10}$. The unit vector in the direction $\displaystyle \overline{\imath} - 3\overline{\jmath}$ is thus $\displaystyle \hat{u} = \frac{1}{\sqrt{10}} (1, -3)$

Directional derivative is then
$\displaystyle D_{\hat{u}} f(a,b) = \hat{u} \cdot \nabla f = \frac{1}{\sqrt{10}} (1,-3) \cdot (4,2)$


$\displaystyle
=\frac{1}{\sqrt{10}} (4-6) = \underline{\underline{\frac{-2}{\sqrt{10}}}}
$

Calculations above are the same no matter when we plugin the point (2,-1).
I think that representing the gradient as a vector is more clear than function.

Now, I will repeat my question:
Is there some cases and/or good reasons why should use functional presentation (i,j components) instead of vector presentation in points (x,y)?