(Vector-valued function derivatives) The Chain Product Property

Let \(\textbf{r}(t)\) be defined as follows:

$$ \textbf{r}(t) = f(t)\textbf{i} + g(t)\textbf{j} + h(t)\textbf{k} $$

The derivative is:

$$ \frac{d}{dt} \textbf{r}(t) = f'(t)\textbf{i} + g'(t)\textbf{j} + h'(t)\textbf{k} $$

If we use \(m(t)\) instead of \(t\):

$$ \frac{d}{d \ m(t)} \textbf{r}(m(t)) = f'(m(t))\textbf{i} + g'(m(t))\textbf{j} + h'(m(t))\textbf{k} $$

If we want to derivative with respect to \(t\), we will have to use the chain rule:

$$\begin{align} \frac{d}{d t} \textbf{r}(m(t)) &= f'(m(t))m'(t)\textbf{i} + g'(m(t))m'(t)\textbf{j} + h'(m(t))m'(t)\textbf{k} \\ &= (f'(m(t))\textbf{i} + g'(m(t))\textbf{j} + h'(m(t))\textbf{k}) m'(t) \end{align}$$

This means:

$$ \frac{d}{d t} \textbf{r}(m(t)) = \textbf{r}'(m(t)) m'(t) $$

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