Differentiation of Vectors, Dot Product, and Cross Product in Cartesian Coordinates
Formulas
Let’s say $\mathbf{A} = A_{x}\hat{\mathbf{x}} + A_{y}\hat{\mathbf{y}} + A_{z}\hat{\mathbf{z}}, \mathbf{B} = B_{x}\hat{\mathbf{x}} + B_{y}\hat{\mathbf{y}} + B_{z}\hat{\mathbf{z}}$ is a vector in a 3-dimensional Cartesian coordinate system. Let $n$ be any scalar. Then, the following equations hold:
(a) $\dfrac{ d \left( n \mathbf{A} \right) }{dt} = \dfrac{ dn }{dt} \mathbf{A} + n\dfrac{ d\mathbf{A}}{dt}$
(b) $\dfrac{ d ( \mathbf{A} \cdot \mathbf{B} )}{dt} = \dfrac{ \mathbf{A} }{dt} \cdot \mathbf{B} + \mathbf{A} \cdot \dfrac{ d\mathbf{B}}{dt}$
(c) $\dfrac{ d ( \mathbf{A} \times \mathbf{B}) }{dt} = \dfrac{ d \mathbf{A} } {dt} \times \mathbf{B} + \mathbf{A} \times \dfrac{ d \mathbf{B} } {dt}$
Explanation
One can naturally accept the results by recalling the product rule of differentiation that is learned since high school. First, let’s calculate the derivative of $\mathbf{A}$.
$$ \begin{align*} \dfrac{ d \mathbf{A} } { dt } =& \dfrac{d}{dt} \left(A_{x}\hat{\mathbf{x}} + A_{y}\hat{\mathbf{y}} + A_{z}\hat{\mathbf{z}} \right) \\ =& \dfrac{ d A_{x} }{dt} \hat{\mathbf{x}} + A_{x} \dfrac{d \hat{\mathbf{x}}}{dt} + \dfrac{ d A_{y} }{dt} \hat{\mathbf{y}} + A_{y} \dfrac{d \hat{\mathbf{y}}}{dt} + \dfrac{ d A_{z} }{dt} \hat{\mathbf{z}} + A_{z} \dfrac{d \hat{\mathbf{z}}}{dt} \\ =& \dfrac{ d A_{x} }{dt} \hat{\mathbf{x}} + \dfrac{ d A_{y} }{dt} \hat{\mathbf{y}} + \dfrac{ d A_{z} }{dt} \hat{\mathbf{z}} \\ =& \left( \dfrac{ d A_{x} }{dt}, \dfrac{ d A_{y} }{dt}, \dfrac{ d A_{z} }{dt} \right) \end{align*} $$
Since the unit vectors in each direction do not change over time, their derivatives are $\mathbf{0}$. From this result, we can see that the derivative of a vector function is still a vector function. Moreover, it can be seen that the following equation also holds.
$$ \left( \dfrac{ d \mathbf{A} } {dt} \right)_{x}=\dfrac{ d A_{x}}{ dt} $$
However, this equation is not generally valid but only holds in a Cartesian coordinate system, so it must be observed with caution. The equation does not hold when unit vectors change over time. For example, in polar coordinates, velocity and acceleration are as follows.
$$ \begin{align*} \mathbf{v}=&\dot{r} \hat{\mathbf{r}} + r \dot{\theta} \hat{\boldsymbol{\theta}} \\ \mathbf{a}=& (\ddot r -r\dot{\theta} ^2)\hat{\mathbf{r}} + (2\dot{r} \dot{\theta} + r\ddot{\theta})\hat{\boldsymbol{\theta}} \end{align*} $$
Then, one can see that the following equality does not hold.
$$ \left( \dfrac{d \mathbf{v}}{dt} \right)_{\theta} = a_{\theta} = 2\dot{r} \dot{\theta} + r\ddot{\theta} \ne \dot{r}\dot{\theta} + r\ddot{\theta} = \dfrac{d v_{\theta}}{dt} $$
Proof
(a)
$$ \begin{align*} \dfrac{ d (n\mathbf{A}) }{ dt} =& \dfrac{ d}{dt} (nA_{x}\hat{\mathbf{x}} + nA_{y}\hat{\mathbf{y}} + nA_{z} \hat{\mathbf{z}}) \\ =& \left( \dfrac{dn}{dt}A_{x} + n\dfrac{dA_{x}}{dt} \right) \hat{\mathbf{x}} +\left( \dfrac{dn}{dt}A_{y} + n \dfrac{dA_{y}}{dt} \right) \hat{\mathbf{y}} + \left( \dfrac{ dn }{ dt } A_{z} + n \dfrac{ d A_{z} }{dt } \right) \hat{\mathbf{z}} \\ =& \dfrac{dn}{dt} \left( A_{x} \hat{\mathbf{x}} + A_{y} \hat{\mathbf{y}} + A_{z} \hat{\mathbf{z}}\right) + n \left( \dfrac{ dA_{x}}{dt}\hat{\mathbf{x}} + \dfrac{dA_{y}}{dt}\hat{\mathbf{y}} + \dfrac{dA_{z}}{dt} \hat{\mathbf{z}} \right) \\ =& \dfrac{ dn } {dt } \mathbf{A} + n\dfrac{ d \mathbf{A} }{ dt } \end{align*} $$
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(b)
$$ \begin{align*} \dfrac{ d (\mathbf{A} \cdot \mathbf{B})} {dt} =& \dfrac{d}{dt}\left( A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z} \right) \\ =& \left( {\color{blue}\dfrac{ d A_{x}}{dt}B_{x}} + A_{x}\dfrac{ d B_{x}}{dt} \right) + \left( {\color{blue}\dfrac{ dA_{y}}{dt}B_{y} } + A_{y}\dfrac{d B_{y}}{dt} \right) + \left( {\color{blue} \dfrac{ d A_{z}}{dt}B_{z} }+ A_{z}\dfrac{dB_{z}}{dt}\right) \\ =& \left( {\color{blue}\dfrac{ d A_{x}}{dt}B_{x} + \dfrac{ dA_{y}}{dt}B_{y} + \dfrac{ d A_{z}}{dt}B_{z} } \right) + \left( A_{x}\dfrac{ d B_{x}}{dt} + A_{y}\dfrac{d B_{y}}{dt} + A_{z}\dfrac{dB_{z}}{dt} \right) \\ =& \left[ \left( \dfrac{ d \mathbf{A}}{dt}\right)_{x}B_{x} + \left(\dfrac{ d\mathbf{A}}{dt}\right)_{y}B_{y} + \left( \dfrac{ d \mathbf{A}}{dt}\right)_{z}B_{z} \right] + \left[ A_{x} \left( \dfrac{ d \mathbf{B}}{dt} \right)_{x} + A_{y} \left( \dfrac{d \mathbf{B}}{dt}\right)_{y} + A_{z} \left( \dfrac{d \mathbf{B} }{dt} \right)_{z}\right] \\ =&\dfrac{d\mathbf{A}}{dt}\cdot \mathbf{B} + \mathbf{A}\cdot \dfrac{d\mathbf{B}}{dt} \end{align*} $$
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(c)
$$ \begin{align*} \dfrac{d( \mathbf{A} \times \mathbf{B}) }{ dt } =& \dfrac{ d}{dt} \left[ \hat{\mathbf{x}} (A_{y}B_{z}-A_{z}B_{y})+ \hat{\mathbf{y}} (A_{z}B_{x}-A_{x}B_{z}) + \hat{\mathbf{z}} (A_{x}B_{y}-A_{y}B_{x}) \right] \\ =& \hat{\mathbf{x}} \left( {\color{blue} \dfrac{ dA_{y}}{dt}B_{z} } +A_{y}\dfrac{d B_{z}}{dt} {\color{blue} -\dfrac{d A_{z}}{dt}B_{y} } –A_{z}\dfrac{B_{y}}{dt} \right) + \hat{\mathbf{y}} \left( {\color{blue} \dfrac{ dA_{z}}{dt}B_{x} } +A_{z}\dfrac{d B_{x}}{dt} {\color{blue} -\dfrac{d A_{x}}{dt}B_{z} } –A_{x}\dfrac{B_{z}}{dt} \right) \\ &\quad + \hat{\mathbf{z}} \left( {\color{blue} \dfrac{ dA_{x}}{dt}B_{y} } +A_{x}\dfrac{d B_{y}}{dt} {\color{blue} -\dfrac{d A_{y}}{dt}B_{x} } –A_{y}\dfrac{B_{x}}{dt} \right) \\ =& {\color{blue} \left[ {\color{black} \hat{\mathbf{x}} \left( \dfrac{ dA_{y}}{dt}B_{z}-\dfrac{d A_{z}}{dt}B_{y}\right) + \hat{\mathbf{y}} \left( \dfrac{ dA_{z}}{dt}B_{x}-\dfrac{d A_{x}}{dt}B_{z} \right) + \hat{\mathbf{z}} \left( \dfrac{ dA_{x}}{dt}B_{y}-\dfrac{d A_{y}}{dt}B_{x} \right)} \right] } \\ &\quad + \left[ \hat{\mathbf{x}} \left( A_{y}\dfrac{d B_{z}}{dt}-A_{z}\dfrac{d B_{y}}{dt} \right) + \hat{\mathbf{y}} \left( A_{z}\dfrac{d B_{x}}{dt}-A_{x}\dfrac{d B_{z}}{dt} \right) + \hat{\mathbf{z}} \left( A_{x}\dfrac{d B_{y}}{dt}-A_{y}\dfrac{d B_{x}}{dt} \right) \right] \\ =& \dfrac{d\mathbf{A}}{dt} \times \mathbf{B} + \mathbf{A} \times \dfrac{d \mathbf{B}}{dt} \end{align*} $$
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