I'm having trouble grasping the use of the wedge product in (what I think is) an exterior derivative. I found this following equation on the wikipedia page (https://en.wikipedia.org/wiki/Exterior_derivative), but am not sure if I was applying it correctly: 
Is this the proper way to solve an exterior derivatives problem, like the first example below? And how would you solve the second part?
Example--
Consider the following differential form fields on $ℝ^3$: $ = ^3 −4x^2 $ and $= ^3 ∧ +sin ∧$
- Find the derivative $$.
I attempted the following, but it doesn't make much sense to me:
$= \frac{∂}{∂x}(z^3) ∧∧ + \frac{∂}{∂y}(z^3) y∧∧ + \frac{∂}{∂x}(sinz) ∧∧ + \frac{∂}{∂z}(sinz) z∧∧z$
so $= 0$..?
$= 0 + 0 + 0+ (cosz)∧∧z$
$= (cosz)∧∧z$
Is that correct? If so what does that result mean?
- Find $∧$.
I'm not sure how to go about solving this part.
$ \def\red#1{\color{red}{#1}} \def\green#1{\color{limegreen}{#1}} \def\blue#1{\color{blue}{#1}} \def\orange#1{\color{orange}{#1}} $
Let $\alpha = x^3dx − 4x^2ydz$ and $\beta= z^3 dx\wedge dy +\sin (z) dx\wedge dz$ as in your question.
First to compute $d\beta$:
$$ \begin{align} d\beta &= d(z^3dx\wedge dy) + d(\sin(z)dx\wedge dz)\\ &=\frac{\partial z^3}{\partial x}dx\wedge dx\wedge dy + \frac{\partial z^3}{\partial y}dy\wedge dx\wedge dy+\frac{\partial z^3}{\partial z}dz\wedge dx\wedge dy\\ &\hspace{.5cm}+\frac{\partial \sin(z)}{\partial x}dx\wedge dx\wedge dz+\frac{\partial \sin(z)}{\partial y}dy\wedge dx\wedge dz+\frac{\partial \sin(z)}{\partial z}dz\wedge dx\wedge dz\\ &=3z^2dz\wedge dx\wedge dy\,,\\ d\beta&= 3z^2 dx\wedge dy\wedge dz\,. \end{align}$$
Now let's look at $\alpha\wedge \beta$: $$\begin{align} \alpha\wedge\beta&=(\red{x^3dx}-\green{4x^2ydz})\wedge(\blue{z^3dx\wedge dy} + \orange{\sin(z)dx\wedge dz})\\ &=\red{x^3dx}\wedge(\blue{z^3dx\wedge dy}+\orange{\sin(z) dx\wedge dz})\\ &\hspace{.5cm}-\green{4x^2ydz}\wedge(\blue{z^3dx\wedge dy}+\orange{\sin(z) dx\wedge dz})\\ &=\red{x^3}\blue{z^3}\red{dx}\wedge \blue{dx\wedge dy} +\red{x^3}\orange{\sin(z)}\red{dx}\wedge\orange{ dx\wedge dz}\\ &\hspace{.5cm}-\green{4x^2y}\blue{z^3}\green{dz}\wedge \blue{dx\wedge dy}+\green{4x^2y}\orange{\sin(z)}\green{dz}\wedge \orange{dx\wedge dz}\\ &=-\green{4x^2y}\blue{z^3dx\wedge dy}\wedge \green{dz} \end{align}$$