# Using the Chain Rule for Vector Calculus: A Tutorial

• I
• binbagsss
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#### binbagsss

TL;DR Summary
chain rule order of differentiation in the product
This is probably a stupid question, but I have never realised that there's an order things should be done in the chain rule , so for example

## \nabla(\bf{v}.\bf{v})=2\bf{v} (\nabla\cdot \bf{v}) ##

and not

## 2 \bf{v} \cdot \nabla \bf{v} ##

Is there an obvious way to see / think of this from the chain rule, say in 1-D, preferably through looking at the limit definition?
Thanks

binbagsss said:
TL;DR Summary: chain rule order of differentiation in the product

and not
It contains gradient of vector which is a tough object.

binbagsss said:
TL;DR Summary: chain rule order of differentiation in the product

This is probably a stupid question, but I have never realised that there's an order things should be done in the chain rule , so for example

## \nabla(\bf{v}.\bf{v})=2\bf{v} (\nabla\cdot \bf{v}) ##

and not

## 2 \bf{v} \cdot \nabla \bf{v} ##

Is there an obvious way to see / think of this from the chain rule, say in 1-D, preferably through looking at the limit definition?
Thanks
The gradient of a scalar function is a vector. All these identities follow from the definition.

topsquark and DaveE
binbagsss said:
TL;DR Summary: chain rule order of differentiation in the product

This is probably a stupid question, but I have never realised that there's an order things should be done in the chain rule , so for example

## \nabla(\bf{v}.\bf{v})=2\bf{v} (\nabla\cdot \bf{v}) ##

and not

## 2 \bf{v} \cdot \nabla \bf{v} ##

Is there an obvious way to see / think of this from the chain rule, say in 1-D, preferably through looking at the limit definition?
Thanks

Using suffix notation, we can form five vectors from two copies of $\mathbf{v}$ and a single $\nabla$: $$\begin{array}{cc} \nabla (\mathbf{v} \cdot \mathbf{v}) & \partial_i(v_jv_j), \\ \nabla \cdot (\mathbf{v} \mathbf{v}) & \partial_j(v_iv_j) , \\ \mathbf{v} \cdot (\nabla \mathbf{v}) & v_j \partial_i v_j, \\ \mathbf{v} (\nabla \cdot \mathbf{v}) & v_i \partial_j v_j, \\ (\mathbf{v} \cdot \nabla) \mathbf{v} & v_j \partial_j v_i. \end{array}$$ Applying the product rule to the first two we have $$\begin{split} \nabla (\mathbf{v} \cdot \mathbf{v}) &= 2\mathbf{v} \cdot (\nabla \mathbf{v}) \\ \nabla \cdot (\mathbf{v} \mathbf{v}) &= (\mathbf{v} \cdot \nabla) \mathbf{v} + \mathbf{v}(\nabla \cdot \mathbf{v}).\end{split}$$ This is about the point at which suffix notation becomes clearer than vector notation.

EDIT: For completeness, we can also form these three using the cross product: $$\begin{array}{cc} \nabla \times (\mathbf{v} \times \mathbf{v}) & \epsilon_{ijk}\epsilon_{klm}\partial_j(v_lv_m) \\ \mathbf{v} \times (\nabla \times \mathbf{v}) & \epsilon_{ijk}\epsilon_{klm} v_j\partial_l v_m \\ (\mathbf{v} \times \nabla) \times \mathbf{v} & -\epsilon_{ijk}\epsilon_{klm} v_l\partial_mv_j \end{array}$$ These can, however, be expressed in terms of the previous vectors by use of the identity $$\epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}.$$

Last edited:
PhDeezNutz, vanhees71 and topsquark
@pasmith, what operation is implied in this product: ##\mathbf{v} \mathbf{v}##? (The 2nd of your 5 examples)

Mark44 said:
@pasmith, what operation is implied in this product: ##\mathbf{v} \mathbf{v}##? (The 2nd of your 5 examples)
Tensor product: $(\mathbf{v}\mathbf{v})_{ij} = v_i v_j$.

vanhees71 and Mark44