## Gradient and directional derivative way of represent the vector equations

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 $(2,1)$ versus using i and j components $2i + j$

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

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

Function presentation:

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

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

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

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

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

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

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

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

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

Vector presentation

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

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

$= (2x,2y)$

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

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

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

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

$
=\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)?