Derivative of Product of Real Function and Vector-Valued Function
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Theorem
Let:
- $\mathbf z:\R \to \R^n$
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 differentiable real functions.
Let:
- $y: \R \to \R$
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
| \(\ds D_x \left({y \, \mathbf{z} }\right)\) | \(=\) | \(\ds D_x \left({\begin{bmatrix} y \ z_1 \\ y \ z_2 \\ \vdots \\ y \ z_n \end{bmatrix} }\right)\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{bmatrix} D_x \left({y \ z_1}\right) \\ D_x \left({y \ z_2 }\right) \\ \vdots \\ D_x \left({y \ z_n }\right) \end{bmatrix}\) | Differentiation of Vector-Valued Function Componentwise | |||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{bmatrix} \dfrac {\d y} {\d x} z_1 + y \dfrac {\d z_1} {\d x} \\ \dfrac {\d y} {\d x}z_2 + y \dfrac {\d z_2} {\d x} \\ \vdots \\ \dfrac {\d y} {\d x} z_n + y \dfrac {\d z_n} {\d x} \end{bmatrix}\) | Product Rule for Derivatives of Real Functions | |||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{bmatrix} \dfrac {\d y} {\d x} z_1 \\ \dfrac {\d y} {\d x} z_2 \\ \vdots \\ \dfrac {\d y} {\d x} z_n \end{bmatrix} + \begin{bmatrix} y \dfrac {\d z_1} {\d x} \\ y \dfrac {\d z_2} {\d x} \\ \vdots \\ y \dfrac {\d z_n} {\d x} \end{bmatrix}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\d y} {\d x} \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix} + y \begin{bmatrix} \dfrac {\d z_1} {\d x} \\ \dfrac {\d z_2} {\d x} \\ \vdots \\ \dfrac {\d z_n} {\d x} \end{bmatrix}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\d y} {\d x} \mathbf z + y \frac {\d \mathbf z} {\d x}\) | Differentiation of Vector-Valued Function Componentwise |
$\blacksquare$