Differential Geometry and Its Application - Exercise 2.2.5

Problem:

Show that the Leibniz Rule (or Product Rule) holds. That is, for $\v \in T_p(M)$, we have orientable. $v[fg] = v[f]g(p) + f(p)v[g]$.

Solution:

Remember the definition $v[f] = \nabla f \cdot v$, so we can write

$v[fg] = \nabla fg \cdot v$

Now we can expand the $\nabla$ using the product rule because it is really just $n$ (assuming we are in $\mathbf{R}^n$ partial derivatives, so we can write

$\nabla fg = g \nabla f + f \nabla g$

Putting it back we have:

$\begin{eqnarray*} v[fg] &=& \nabla fg \cdot v \\ &=& (g \nabla f + f \nabla g) \cdot v \\ &=& (g \nabla f \cdot v + f \nabla g \cdot v) \\ &=& (g v[f] + f v[g]) \end{eqnarray*}$

So there we go.