Derivative of Product of Real Function and Vector-Valued Function

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Theorem

Let:

$\mathbf r:x \mapsto \mathbf z$

be a differentiable vector-valued function, where:

$\mathbf{z} = \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix}$

such that:

$z_1, z_2, \cdots, z_n$

are (images of) differentiable real functions.

Let:

$f: x \mapsto y$

be a differentiable real function.


Then:

$D_x \left({y \, \mathbf z}\right) = \dfrac {\d y} {\d x} \mathbf z + y \dfrac {\d \mathbf z} {\d x}$


Proof

\(\displaystyle D_x \left({y \, \mathbf{z} }\right)\) \(=\) \(\displaystyle D_x \left({\begin{bmatrix} y \ z_1 \\ y \ z_2 \\ \vdots \\ y \ z_n \end{bmatrix} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} D_x \left({y \ z_1}\right) \\ D_x \left({y \ z_2 }\right) \\ \vdots \\ D_x \left({y \ z_n }\right) \end{bmatrix}\) Differentiation of Vector-Valued Function Componentwise
\(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} \dfrac {\d y} {\d x} z_1 + y \dfrac {\d z_1} {\d x} \\ \dfrac {\d y} {\d x}z_2 + y \dfrac {\d z_2} {\d x} \\ \vdots \\ \dfrac {\d y} {\d x} z_n + y \dfrac {\d z_n} {\d x} \end{bmatrix}\) Product Rule for Derivatives of Real Functions
\(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} \dfrac {\d y} {\d x} z_1 \\ \dfrac {\d y} {\d x} z_2 \\ \vdots \\ \dfrac {\d y} {\d x} z_n \end{bmatrix} + \begin{bmatrix} y \dfrac {\d z_1} {\d x} \\ y \dfrac {\d z_2} {\d x} \\ \vdots \\ y \dfrac {\d z_n} {\d x} \end{bmatrix}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\d y} {\d x} \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix} + y \begin{bmatrix} \dfrac {\d z_1} {\d x} \\ \dfrac {\d z_2} {\d x} \\ \vdots \\ \dfrac {\d z_n} {\d x} \end{bmatrix}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\d y} {\d x} \mathbf z + y \frac {\d \mathbf z} {\d x}\) Differentiation of Vector-Valued Function Componentwise

$\blacksquare$


Also see