d(dx) exterior derivative in simple language? I’m only learning the basics of analysis, so I have no understanding of exterior derivatives. I simply need clarification on my misconception using simple words.
I’m reading about second degree derivatives and found the following
$$d\left( dy\right) =d\left( f'\left( x\right) dx\right)= f''(x) dx * dx= f''\left( x\right) \left( dx\right) ^{2}$$
So I was thinking of expanding $d(f’(x)dx)$ to $f’’(x) * dx + f’(x) * d(dx)$. According to the axioms of exterior derivative, the latter part should equals to zero. However I was thinking $dx = 1 * \triangle(x)=(x-x_0)$, $d(dx) = d(x - x_0) = d(x) - d(x_0) = d(x)$
According to the definition of the exterior derivative, this is wrong. I’m confused with the misconception I have. Please help me clarify this.
Update on maybe a partial solution,
I’ve done some research on the web and found that maybe dx should be viewed as a constant. In that case d(dx) = 0. Can we view dx as a constant at all?
 A: I would suggest you use the notation $D$ instead of $d$, because $d$ refers to the exterior derivative, so that for any $C^2$ function $f$, we ALWAYS have $d^2f:=d(df)=0$. This is one of the standard theorems when learning about exterior derivatives.
What I think you're after is the second (and presumably higher order Frechet) derivatives $Df,D^2f,D^3f,\dots D^k f$. In this case, if $f:\Bbb{R}^n\to\Bbb{R}$ is a sufficiently smooth function, then $D^kf$ is a tensor field of type $(0,k)$ on $\Bbb{R}^n$, so if we let $\{e_1,\dots, e_n\}$ be the standard ordered basis for $\Bbb{R}^n$ then
\begin{align}
D^kf&=\sum_{I}(D^kf)[e_{i_1},\dots, e_{i_k}]\,dx^{i_1}\otimes\cdots \otimes dx^{i_k}\\
&=\sum_{I}\frac{\partial^kf}{\partial x^{i_1}\cdots \partial x^{i_k}}\,dx^{i_1}\otimes\cdots \otimes dx^{i_k}\\
&\equiv \sum_{I}\frac{\partial^kf}{\partial x^I}\,dx^{\otimes I}
\end{align}
(the first equality is really a matter of linear algebra, regarding how we can always expand a tensor in terms of a given basis), and in the last equal sign, I have just used shorthand notation.
Anyway, all this equality is saying is that at any given point $p\in\Bbb{R}^n$, $(D^kf)_p$ is a $(0,k)$-tensor on $\Bbb{R}^n$ whose value on a $k$-tuple of vectors $(h_1,\dots, h_k)\in\underbrace{\Bbb{R}^n\times\cdots \times \Bbb{R}^n}_{\text{$k$ times}}$ is
\begin{align}
(D^kf)_p[h_1,\dots, h_k]&=\sum_{I}\frac{\partial^k f}{\partial x^I}(p)\cdot (h_1)^{i_1}\cdots (h_k)^{i_k}.
\end{align}
Of course, when the domain of $f$ is simply $\Bbb{R}$ (i.e $n=1$) then the linear algebra is particularly simple, because the tensor $D^kf$ is determined completely determined by the $k^{th}$ derivative $f^{(k)}$ in the sense that
\begin{align}
D^kf&=f^{(k)}\,dx^{\otimes k}
\end{align}
In this case, one may prefer to drop the tensor product symbol and simply write
\begin{align}
D^kf&=f^{(k)}\,(dx)^k
\end{align}
Once again, the meaning of this notation is that if you take derivatives at a point $p\in \Bbb{R}$ and evaluate on $(h_1,\dots, h_k)\in\underbrace{\Bbb{R}\times\cdots\times\Bbb{R}}_{\text{$k$ times}}$, then we get
\begin{align}
(D^kf)_p[h_1,\dots, h_k]&=f^{(k)}(p)\cdot h_1\cdots h_k.
\end{align}

Calculus is often pretty funny because we often introduce a lot of terminology:

*

*The object $Df$ is often called the total derivative (in this case it just so happens that by identifying all the tangent spaces this equals the exterior derivative $df$).

*The object $D^2f$ is often called the Hessian of $f$ and is denoted as $H_f$ (though people often think of the Hessian as a certain $n\times n$ matrix... but this only works if the target space of $f$ is $\Bbb{R}$. If we change the target space to a higher dimensional space, then one is forced to abandon matrices and use the full force of multilinear algebra/tensors).

