let $\alpha\in\Omega^1(U, \mathbb{R})$ be a 1-form on $U$ and $X, Y : U → \mathbb{R}^n$ vector fields. Verify that the exterior derivative can be calculated via $$d\alpha(X, Y ) = X \cdot\alpha(Y ) − Y \cdot \alpha(X) − \alpha([X, Y ])$$ where the smooth function $\alpha(X): U → \mathbb{R}$ is given by $$\alpha(X)(p) := \alpha_p(X(p))$$ for $p\in U$.
I haven't seen a ton on the exterior derivative and the bit that I have has involved sort of how it operates in the abstract on the cochain complex. In other words showing how it moves a 1-form to a 2 form and so on..
My instinct would be to rewrite the equation for the exterior derivative using the definition of the function $\alpha$(X) and the definition of the lie bracket and see where that leads me. Or alternatively, to rewrite the equation in order to equate the lie bracket(multiplied by $\alpha$ (the one form)) to the remaining terms and hope that they work out nicely. Any help?
Since $d\alpha$ is a differential form, and in particular a tensor, the value of $d\alpha(X,Y)$ at the point $p$ only depends on the values of $X$ and $Y$ at $p$. Hence, you may assume that $X$ and $Y$ are constant vector fields. This makes the whole computation much simpler.