# How does the gradient/jacobian relate to the first derivative of one dimensional functions?

My question is about this answer by Simon S (Is the gradient vector of a function the derivative of the function)

The derivative of a function $f : U \subset \mathbb{R}^m \to \mathbb{R}$ at a point $p \in U$ is a linear map $Df_p : \mathbb{R}^m \to \mathbb{R}$. We can identify the gradient of $f$ with $D_p$, if everything exists, by

$$Df_p(v) = \langle \nabla f(p), v\rangle$$

Equivalently, $Df_p$ is a co-tangent vector, i.e., a member of the set of linear maps acting on $T_p$, the vector space of tangent vectors at $p$: i.e., $Df_p = \langle \nabla f(p), \cdot \rangle : T_p \to \mathbb{R}$. This formalism is helpful when we want to generalize from $\mathbb{R}^n$ to manifolds.

In an abstract sense, I think I understand the differential being a linear map. But in $\mathbb{R}$ we just have one derivative $f'(x)$ where we plug in a value for x and get a scalar that tells us the derivative at that point.

But in higher dimensions, we establish a gradient or a Jacobian at a point and then we use the dot product to calculate the derivative as stated above. Of what importance is it at which point we create the gradient and of what importance which vector we multiply with the gradient?

Is $\nabla f(3,3)$ the derivative of the function $f:\mathbb{R^2}\rightarrow \mathbb{R}$? Or is it $\langle\nabla f(x,y),(3,3)\rangle$ Or is it $\langle\nabla f(3,3),(3,3)\rangle$ ? How can one understand this as just the higher dimensional derivative comparable to a single dimensional $f'(3)$? Why do we need that dot product with an additional vector?

Where is the connection between the derivative being a linear map and the derivative just being the sum of the partial derivatives?

EDIT: If it isn't clear, what I'm implying consider sites like these http://www.solitaryroad.com/c353.html or https://en.wikipedia.org/wiki/Total_derivative which essentially state

The total differential is the sum of the partial differentials.

But I can not fathom how that is equivalent to the linear map definition above.

Stick to the idea that the derivative of an ${\bf f}:\>{\mathbb R}^n\to{\mathbb R}^m$ at a point ${\bf p}\in{\rm dom}({\bf f})$ is a linear map $$L:=D{\bf f}({\bf p}):\>{\mathbb R}^n\to{\mathbb R}^m\ .$$ This map is uniquely defined by the condition $${\bf f}({\bf p}+{\bf X})-{\bf f}({\bf p})=L\>{\bf X}+o\bigl(|{\bf X}|\bigr)\qquad({\bf X}\to{\bf 0})\ .\tag{1}$$ This holds also when $m=1$ and/or $n=1$. In the case $m=1$ ($f$ a multivariate scalar function) the map $L$ is a linear functional $L:\>T_{\bf p}\to{\mathbb R}$. If a scalar product is present then this functional $L$ is represented by a vector ${\bf a}$ in the form $L\>{\bf X}={\bf a}\cdot{\bf X}$. This vector ${\bf a}$ is called the gradient of $f$ at ${\bf p}$; its components are the partial derivatives of $f$ at ${\bf p}$: $$\nabla f({\bf p})=\left({\partial f\over\partial x_1},\ldots,{\partial f\over\partial x_n}\right)_{\bf p}\ .$$ The formula $(1)$ then appears as $$f({\bf p}+{\bf X})-f({\bf p})=\nabla f({\bf p})\cdot {\bf X}+o\bigl(|{\bf X}|\bigr)\qquad({\bf X}\to{\bf 0})\ .$$ If we write this out in coordinates we get $$f(p_1+X_1,\ldots, p_n+X_n)-f(p_1,\ldots, p_n)=\sum_{k=1}^n f_{.k}({\bf p})\>X_k+o\bigl(|{\bf X}|\bigr)\qquad({\bf X}\to{\bf 0})\ .$$ In particular one has $$Df(p_1,\ldots, p_n).(X_1,\ldots X_n)=\sum_{k=1}^n f_{.k}({\bf p})\>X_k\ .$$ The statement that "$Df({\bf p})$ is just the sum of the partial derivatives of $f$ at ${\bf p}$" is pure nonsense.