units of terms in partial derivatives

Question

Consider: the function f is a function of u, which is a function of a and b. $$ \frac{\partial^2f}{\partial a \partial b} = \frac{\partial}{\partial b}(\frac{df}{du}\frac{\partial u}{\partial a}) $$ $$ \frac{\partial^2f}{\partial a \partial b} = \frac{\partial u}{\partial a}\frac{\partial}{\partial b}(\frac{df}{du})+\frac{df}{du}\frac{\partial^2u}{\partial a \partial b} $$ $$ \frac{\partial^2f}{\partial a \partial b} = \frac{\partial u}{\partial a}\frac{d^2f}{du^2}\frac{\partial u}{\partial b}+\frac{df}{du}\frac{\partial^2u}{\partial a \partial b} $$ The units on the left hand side is [f^2]/([a][b]). I'm struggling to reconcile that with the units of the second term on the right hand side, which is [f][u]/([a][b]). Thanks in advance.

Answer 1

Think about position $x(t)$, velocity $v(t)=\dfrac{dx}{dt}$, and acceleration $a(t) = \dfrac{d^2x}{dt^2}$. The units are ft (or m), ft/sec, and ft/sec$^2$. The second derivative is rate of change of the first derivative, so that extra rate of change contributes another /sec in the denominator, but the units of ft do not change.

Thus, in your analysis, both sides have units of $[f]/[a][b]$. On the right-hand side, you have $[f]/[u]^2\cdot [u]/[a]\cdot [u]/[b]$ and $[f]/[u]\cdot [u]/[a][b]$, respectively.