Derivative function $\frac 1f$ in the sense of distributions

Question

$\newcommand\interval{\mathopen]0,T\mathclose[}$Let $f \in W^{1,\infty }(\interval)$ and $f(x)>1$ a.e. $x\in \interval$. Then $g = \frac 1f \in W^{1,\infty}(\interval)$. (g is Lipschitz in $\interval$.)

I want to show it again by use derivative of g.

Is it reasonable to say that $g'= \frac{-f '} {f^2}$ in $D'$?. If yes, I need a way to prove it

Answer 1

For $\phi \in C^\infty_c,\phi \ge 0, \int\phi=1$.

$F_n=F\ast n \phi(nx)$ is $C^\infty$ and it converges to $F$ locally uniformly.

Since $F>1$ then $\frac{1}{F_n}$ converges to $\frac1{F}$ uniformly.

And $\frac{-F_n'}{F_n^2}$ converges to $\frac{-F'}{F^2}$ in $D'$.

ie. $\frac{-F'}{F^2}$ is the distributional derivative of $\frac1{F}$.

In $D'$: $\frac{-F'}{F^2}=\frac{-f'}{f^2},\frac1{F}=\frac1f$.