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Is $r \times \frac{d}{dt} mv=\frac{d}{dt} (r \times mv)$

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Is $$r \times \frac{d}{dt} mv=\frac{d}{dt} (r \times mv)$$

I was thinking that it is wrong. Cause, which is outside of differentiation how we can put that inside differentiation no matter that's constant or not.

1.9 differentiation

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What is $\vec{v}\times\vec{v}$? (2 comments)

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In equations 1.9 and 1.10, the quantities $\mathbf{r}$, $\mathbf{F}$, $\mathbf{N}$, and $\mathbf{v}$ are vectors, and the symbol $\times$ denotes a vector cross product operation.

The product rule of differential calculus also applies to the vector cross product.

That is, $\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{a}\times \mathbf{b})=\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{a})\times\mathbf{b}+\mathbf{a}\times\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{b})$.

Thus,

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{r}\times m\mathbf{v}) &=\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{r})\times m\mathbf{v}+\mathbf{r}\times\frac{\mathrm{d}}{\mathrm{d}t}(m\mathbf{v})\newline &=\mathbf{v}\times m\mathbf{v}+\mathbf{r}\times\frac{\mathrm{d}}{\mathrm{d}t}(m\mathbf{v})\newline \end{align*}

Since $\mathbf{v}\times m\mathbf{v}=m(\mathbf{v}\times\mathbf{v})=m\mathbf{0}=\mathbf{0}$, we have \[\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{r}\times m\mathbf{v})=\mathbf{r}\times\frac{\mathrm{d}}{\mathrm{d}t}(m\mathbf{v})\]

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