Do you see why:
$m\int \frac{dv}{dt} v dt = \frac{m}{2}\int \frac{d}{dt}(v)^2 dt = $
How can you put the $v$ inside the differential $\frac{d}{dt}$, if it is time-dependent? And where does the $\frac{1}{2}$ come from?
Another (easy) one:
$\mathbf r \times \mathbf (\frac{d}{dt}(mv))=\frac{d}{dt}(\mathbf r \times \mathbf mv)$
How can you bring the r into the differential? Must r be time independent for that?
Thank you very much
The second one uses $$\frac{d}{dt}\left(\vec{r}\times\vec{p}\right)=\frac{d\vec{r}}{dt}\times\vec{p}+\vec{r}\times\frac{d\vec{p}}{dt},$$with $\vec{p}:=m\frac{d\vec{r}}{dt}$. The first cross product on the right is therefore of parallel vectors, and vanishes. The second one succumbs to the same logic provided $\frac{d\vec{p}}{dt}$ - i.e. the force in Newton's second law - is parallel to $\vec{r}$, and in that case we call the force radial. (For example, if our origin is the Sun, the gravitational force it exerts on a position-$\vec{r}$ Earth is radial).
For the first one, by the chain rule, $$\frac{d}{dt}(v^2) = (2v) \frac{dv}{dt}$$ and if we divide both sides by $2$ and rearrange slightly: $$\frac{1}{2} \cdot \frac{d}{dt}(v^2) = \frac{dv}{dt}v$$
My explanation for the second one was wrong, so I've removed it
As comment pointed out by Botond, in our case $\bf r$ is not independent of time so the considerations below are not really relevant to this question.
$\bf r$ is independent of time. First of all a cross product is a product. Whenever you have a product of two things and you want to differentiate it we need to consider the multiplication rule for differentials :
$$\frac{\partial(f\cdot g)}{\partial t} = f'g + fg'$$
But, if $g$ is independent of $t$, then for all purposes it is a constant function of $t$. And all constant functions become $0$ when we differentiate. So $g' = 0$ so it basically says:
$$f'g + f\cdot 0 = f'g$$
In our case $\bf r$ is $g$ as it is independent of the time variable.
