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.