In a proof, my professor shows:
$ s = g^{-1}(u) $
$ ds = \frac{dg^{-1}(u)}{du} du $ , by the chain rule
If I were to apply the chain rule to calculate ds, I would not get the du in the denominator. Can someone explain where the du in the denominator comes from?
If you apply thde differential operator $\frac d{du}$ to $s(u)=g^{-1}(u)$, you obtain $\frac{ds}{du}=\frac{dg^{-1}(u)}{du}$ with no chain rule involved. If you transform $s(u)=g^{-1}(u)$ to $g(s(u))=u$, the chain rule produces $g'(s(u))s'(u)=1$.