# Derivative of momentum with respect to velocity

Why the derivative of momentum with respect to velocity equals $$m^3$$? My textbook says: $$d\vec{p}=dp_{x}\cdot dp_{y}\cdot dp_{z}$$ and $$d\vec{u}=du_{x}\cdot du_{y}\cdot du_{z}$$ Then divides the two relations and gets: $$\frac{d\vec{p}}{d\vec{u}}=m^3$$ But I don't understand why the above relations are true. I find in Wikipedia that the derivative of a vector with respect to a vector is given by the matrix:
$$\frac{\partial\vec{y}}{\partial\vec{x}}=\begin{bmatrix} \frac{\partial{y_1}}{\partial{x_1}} & \frac{\partial{y_1}}{\partial{x_1}} & \cdots & \frac{\partial{y_1}}{\partial{x_n}}\\ \frac{\partial{y_2}}{\partial{x_1}} & \frac{\partial{y_2}}{\partial{x_2}} & \cdots & \frac{\partial{y_2}}{\partial{x_1}}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{\partial{y_m}}{\partial{x_1}} & \frac{\partial{y_1}}{\partial{x_2}} & \cdots & \frac{\partial{y_m}}{\partial{x_n}} \end{bmatrix}$$ Following this procedure for momentum $$\vec{p}(\vec{u})$$ one finds that: $$\frac{\partial\vec{p}}{\partial\vec{u}} = m\begin{bmatrix} I\\ \end{bmatrix}$$ The only way this to be equal to $$m^3$$ is to take the determinant of the matrix. But, Wikipedia doesn't refer this. Any ideas on how to treat this problem? Thanks in advance.

$$d\vec p$$ is not actually a vector, but a differential volume element. It is a bit of sloppy notation because sometimes $$d\vec p$$ means the vector $$(dp_x,dp_y,dp_z)^T$$ and sometimes it is the volume element $$dp_xdp_ydp_z$$. You can also use $$d^3p$$ to denote a volume element but both are used often.
A more proper way to derive $$m^3$$ is using the Jacobian. You probably remember from multivariable calculus that if you transform multiple coordinates under an integral you have to multiply by the Jacobian: $$\int dp_xdp_ydp_z=\int J(\vec p,\vec u)du_xdu_ydu_z$$ We are only interested in the volume element so you can ignore the integral sign. $$J$$ is given by the determinant of Jacobian matrix*, which happens to be the vector derivative you mention. $$J=det\left(\frac{\partial\vec p}{\partial\vec u}\right)=det\left(m\mathbf I\right)=m^3$$ Taking the determinant of a diagonal matrix is easy: you just multiply all the diagonal component to get $$m^3$$. So the trick the author mentioned only works because the Jacobian is diagonal, or equivalently because the i'th component of $$p$$ only depends on the i'th component of $$u$$. So finally we get $$dp_xdp_ydp_z=m^3du_xdu_ydu_z$$
*To avoid confusion: when I say just the Jacobian I mean the determinant of the Jacobian which is a number and when I say Jacobian matrix I mean the 3x3 matrix $$\partial\vec p/\partial\vec u$$.