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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 $$\frac{\phi(a+h)-\phi(a)}{h} = \frac{g(a)-g(a+h)}{hg(a)g(a+h)}.$$

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up vote 1 down vote accepted

$$ \lim_{h \to 0} \frac{g(a)-g(a+h)}{hg(a)g(a+h)}= \lim_{h \to 0} \left( \frac{g(a)-g(a+h)}{h} \cdot \frac{1}{g(a)g(a+h)} \right) = \\ = -\lim_{h \to 0} \frac{g(a+h)-g(a)}{h} \cdot \lim_{h \to 0} \frac{1}{g(a)g(a+h)} = -g'(a) \cdot \frac{1}{[g(a)]^2} = -\frac{g'(a)}{[g(a)]^2}. $$ Can you recognize the limit of the left hand side if $h$ tends to $0$?

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The left hand side is of course the difference quotient for $\phi$, its limit as $h\to 0$ is $\phi'(a)$. The right hand side can be recognized as $$ -\frac1{g(a)g(a+h)}\cdot\frac{g(a+h)g(a)}{h}$$ As $h\to 0$, the second factor converges to $g'(a)$, and the $g(a+h)$ in the first denominator converges to $g(a)$.

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Great! Just bit stuck of why the second factor is g'(a)? – user127700 Apr 17 '14 at 16:13
I think you're missed minus in the second fraction. – user35603 Apr 17 '14 at 16:19

Filling in the stages from the LHS of the equation stated in the proof $$\frac{\phi(a+h)-\phi(a)}{h}=\frac{\frac{1}{g(a+h)}-\frac{1}{g(a)}}{h}=\frac{g(a)-g(a+h)}{hg(a)g(a+h)}\\=-\left[\frac{g(a+h)-g(a)}{h}\right]\frac{1}{g(a)g(a+h)}$$

If we let $h\rightarrow 0$, the expression becomes $$\lim_{h\rightarrow0}\left\{-\left[\frac{g(a+h)-g(a)}{h}\right]\frac{1}{g(a)g(a+h)}\right\}=-\frac{g'(a)}{g(a)^2}$$

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