relationship between derivative of vector valued function and gradient/partial derivatives From This article I understand that if I have a function defined using parametric equations like this
$$
(1) \quad \quad \quad{{\mathbf{r}\left( t \right) = f\left( t \right)\mathbf{i} + g\left( t \right)\mathbf{j} + h\left( t \right)\mathbf{k}}\;\;\text{ or }\;\;}\kern0pt{\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle }
$$
Then it's derivative is the derivative of each function that generates each coordinate.
$$
(2) \quad \quad \quad {\mathbf{r}^\prime\left( t \right) = \left\langle {f^\prime\left( t \right),g^\prime\left( t \right),h^\prime\left( t \right)} \right\rangle.}
$$
However if the same function is represented in a non parametric form (random example below) 
$$
(3) \quad \quad \quad r( x,y)= x+y  
$$
 It seems there is no such thing as "the derivative", I only have partial derivatives, directional derivatives and the gradient.
What is the relationship between "the" derivative as defined in $(2)$ and the gradient or the partial/directional derivatives? Are they equivalent in some way?
 A: The parametrically defined function
$$r(t) = \begin{pmatrix}f(t)\\g(t)\\h(t)\end{pmatrix}$$
is a function from the real number line $\mathbb R$ to the three dimensional space $\mathbb R^3$. Thus, we can take the standard one-dimensional derivative. The relationship to the directional derivative is that a direction is already chosen by the parametrisation. The directional derivative of a function $f:\mathbb R^n \to\mathbb R^m$ along a direction $v\in\mathbb R^n$ at a point $x_0$ is
$$
\partial_vf(x_0) = \lim_{h\to 0} \frac{f(x_0+hv)-f(x_0)}{h}.
$$
Since the parametrisation is a one-dimensional function, the only possible directions are multiples of $-1$ and $1$. Taking the directional derivative along these directions actually is equivalent to the left-sided and right-sided derivative. If $f$ is differentiable they both agree. For $r$ this is
$$
r'(t) = \begin{pmatrix}f'(t)\\g'(t)\\h'(t)\end{pmatrix}.
$$
The partial derivatives are actually a special case of directional derivatives. They are the directional derivatives along the coordinate axes. That is to say, for your example $g(x,y) = x+y$, the partial derivatives are
$$\frac{\partial}{\partial x}g(x,y) = y\quad\text{and}\quad\frac{\partial}{\partial y}g(x,y)=x.$$
The one-dimensional derivative can be generalised for functions $f\colon\mathbb R^n\to\mathbb R^m$ to the so-called total derivative. For $m=1$ this actually is the transpose of the gradient. The function $f$ is called total differntiable in $x_0$, iff
$$\lim_{x\to x_0}\frac{\lVert f(x)-f(x_0) - Df(x_0)(x-x_0)\rVert}{\lVert x-x_0\rVert},$$
where $Df(x_0)\colon \mathbb R^n\to\mathbb R^m$ is a linear map, exists. Then we call $Df(x_0)$ the total derivative of $f$ at $x_0$. Notice, the similarity to the definition of the one-dimensional derivative. If all partial derivatives exists and are continuous, then the total derivative can be expressed as
$$
Df(x_0) = \biggl(\frac{\partial f}{\partial x_1},\dots, \frac{\partial f}{\partial x_n}\biggr)\in\mathbb R^{m\times n}\,.
$$
Multiplying the total derivative $Df(x_0)$ with a direction gives the directional derivative.
