Why does the result follow? How does this theorem follow?

Theorem. If $g$ is differentiable at $a$ and $g(a) \neq 0$, then $\phi = 1/g$ is also differentiable at $a$, and $$\phi'(a) = (1/g)'(a) = -\frac{g'(a)}{[g(a)]^2}.$$
Proof. The result follows from $$\frac{\phi(a+h)-\phi(a)}{h} = \frac{g(a)-g(a+h)}{hg(a)g(a+h)}.$$


The results follow because:


*

*if $\lim\limits_{h\to 0}\dfrac{\phi(a+h)-\phi(a)}{h}=\ell$ exists ($\in\mathbb{R}$) then $\ell=\phi'(a)$.

*Calculate the limit of this ratio $\dfrac{\phi(a+h)-\phi(a)}{h}=\dfrac{g(a)-g(a+h)}{h}\cdot \dfrac{1}{g(a)g(a+h)}$ knowing that $g$ is defirentiable at $a$.

*You find that $\lim\limits_{h\to 0}\dfrac{\phi(a+h)-\phi(a)}{h}=\phi'(a)=\lim\limits_{h\to 0}\dfrac{g(a)-g(a+h)}{h}\cdot \dfrac{1}{g(a)g(a+h)}=\dfrac{-g'(a)}{g^2(a)}$.

