O'Neill's Elementary Differential Geometry, Exercise 4.6.1, has me mystified. The exercise:
If $\alpha$ is a curve in $R^2$ and $\phi$ is a 1-form, prove this computational rule for finding $\phi(\alpha^\prime)dt$: Substitute $u=\alpha_1$ and $v=\alpha_2$ into the coordinate expression $\phi=f(u,v)du+g(u,v)dv$.
My attempt:
$\phi(\alpha^\prime)=f(u,v)du(\alpha^\prime)+g(u,v)dv(\alpha^\prime)=f(\alpha)\alpha_1^\prime+g(\alpha)\alpha_2^\prime$
I do not see what there is to prove.
What O'Neill means by "Substitute $u=\alpha_1$ and $v=\alpha_2$" is exactly that - every time you see a $u$ in that coordinate expression, replace that with $\alpha_1$. In particular, you should replace the $u$ in $du$ as well. So the "computational rule" is intended to be something like $$\phi(\alpha')dt=f(\alpha_1,\alpha_2)d(\alpha_1) + g(\alpha_1,\alpha_2)d(\alpha_2)$$ There is indeed not much to prove: $$\begin{align} \phi(\alpha')dt &= (fdu + gdv)(\alpha')dt \\ &= (f(\alpha_1,\alpha_2)\alpha'_1 + g(\alpha_1, \alpha_2)\alpha'_2)dt\\ &= f(\alpha_1,\alpha_2)\alpha'_1dt + g(\alpha_1,\alpha_2)\alpha'_2dt \\ &= f(\alpha_1,\alpha_2)d(\alpha_1) + g(\alpha_1,\alpha_2)d(\alpha_2) \,\square \\ \end{align} $$