Action of gradient on tangent vector gives total derivative? In "Spacetime and geometry" Carroll introduces provides the tangent vector (to a curve) as an example of a vector, and the gradient (of a scalar field) as an example of a dual-vector. He then goes on to state:

Note the gradient does in fact act in a natural way on the example we
gave above of a vector, the tangent vector to a curve. The result is
an ordinary derivative of the function along the curve.
$$\partial_\mu\phi\frac{\partial x^\mu}{\partial \lambda} = \frac{d\phi}{d\lambda}$$

(This is on page 20 of my edition, and the equation is 1.55.)
How is the RHS of the equality found? Is it just notation, or has he introduced a rule for combining vectors and dual vectors that goes from partials to total derivatives?
 A: Recall that a parameterized curve through spacetime is specified by the coordinates, $x=(x^0,x^1,x^2,x^3)$, as a function of a parameter, for example, $x^{\mu}(\lambda)$. A tangent vector to this curve, V, is written as:
\begin{equation}
V= V^{\alpha}\hat{e}_{(\alpha)} = \frac{dx^{\alpha}}{d\lambda} \; \hat{e}_{(\alpha)} 
\end{equation}
where $\hat{e}_{(\alpha)}$ are the basis vectors and the dependence of the spacetime coordinates on $\lambda$ is omitted. Recall also that the gradient of a scalar function, $\phi = \phi(x(\lambda))$, is the set of partial derivatives with respect to $x^{\mu}$ taken along some basis vectors $\hat{\theta}^{(\mu)}$:
\begin{equation}
d\phi= \frac{\partial \phi}{\partial x^{\mu}} \hat{\theta}^{(\mu)}= \partial_{\mu}\phi\; \hat{\theta}^{(\mu)}
\end{equation}
If $d\phi$ is a dual vector to V then, the basis vectors are orthonormal:
\begin{equation}
\hat{\theta}^{(\mu)} \; \hat{e}_{(\alpha)} = \delta_{\alpha}^{\mu}
\end{equation}
Thus, we see that the action of $d\phi$ on V is:
\begin{align*}
d\phi\;V &= \partial_{\mu}\phi\; \hat{\theta}^{(\mu)} \;V^{\alpha} \; \hat{e}_{(\alpha)} \\
&\quad= \partial_{\mu}\phi \;V^{\alpha} \;\hat{\theta}^{(\mu)} \hat{e}_{(\alpha)}  \\
&\quad= \partial_{\mu}\phi \;V^{\alpha} \; \delta_{\alpha}^{\mu} \\
&\quad= \partial_{\mu}\phi \;V^{\mu} \\
&\quad= \frac{\partial\phi}{\partial x^{\mu}} \;\frac{dx^{\mu}}{d\lambda} \\
\end{align*}
This is how we obtain the LHS of expression (1.55) of the mentioned book.
To see that
\begin{equation}
 \frac{\partial\phi}{\partial x^{\mu}} \;\frac{dx^{\mu}}{d\lambda} = \frac{d\phi}{d\lambda} 
\end{equation}
just apply the chain rule to the RHS writting the explicit dependence of $\phi$ on $x(\lambda)$
\begin{equation}
\frac{d\phi(x(\lambda))}{d\lambda} = \frac{\partial\phi (x(\lambda))}{\partial x^{\mu}(\lambda)} \;\frac{dx^{\mu}(\lambda)}{d\lambda}
\end{equation}
