text: Alternative proof for differentiability of inverse function?
  
  Problem Let $f$ be a continuous one-one function defined on an interval and suppose that $f$ is differentiable at $f^{-1}(b)$ with the derivative $f'(f^{-1}(x)) \neq 0$. Prove that $f^{-1}(x)$ is also differentiable at $b$.
  
  I read the proof in Spivak Calculus book, but it's quite confusing to me. Then I looked up online, and I saw exact the same proof everywhere. So I wonder is there an alternative proof to this problem that I'm not aware of?
  
  Here is the proof from Spivak book
  
  Theorem 5 (Calculus by Spivak 4th edition, page 237-238)
  Proof. Let $b = f(a)$. Then $$\displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - f^{-1}(b)}{h} = \displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - a}{h}$$ Now every number $b + h$ in the domain of $f^{-1}$ can be written in the form $$b + h = f(a + k)$$ for a unique $k$. Then $$\displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - a}{h} = \displaystyle\lim_{h\to 0}\dfrac{f^{-1}(f(a + k)) - a}{f(a + k) - b} = \displaystyle\lim_{h\to 0} \dfrac{k}{f(a + k) - f(a)}$$ We have $$f^{-1}(b + h) = a + k \Leftrightarrow k = f^{-1}(b + h) - f^{-1}(b)$$ Since $f$ is continuous and one-one, $f^{-1}$ is also continuous at $b$. This means that $k$ approaches $0$ as $h$ approaches $0$. Since $$\displaystyle\lim_{k\to 0} \dfrac{f(a + k) - f(a)}{k} = f'(a) = f'(f^{-1}(b)) \neq 0$$ this implies that $$(f^{-1})'(b) = \dfrac{1}{f'(f^{-1}(b))}$$
  
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  What do you mean by "$f(x)$ is differentiable"? Is that supposed to be "$f$ is differentiable at $f^{-1}(x)$"? – wj32 Dec 5 '12 at 8:41
  @wj32: Yes, I was a little sloppy since I thought it was very known proof. Let me fix that. – Chan Dec 5 '12 at 8:43
  I delete my comments since you edited. – Willie Wong Dec 5 '12 at 9:34
  
  The strategy of the proof is simple, and the difficulty lies in the existence of the limit for the incremental ration of $f^{-1}$. Indeed, if $f^{-1}$ is differentiable, then $f^{-1} \circ f = \operatorname{Id}$ implies $$Df^{-1}(f(x))Df(x)=1$$ for every $x$ in the domain of $f$, and you conclude.
  On the other hand, the existence of the derivative of $f^{-1}$ is a standard application of the change-of-variable rule for limits, which is in turn a corollary of the theorem on limits of composite functions.
explanation: The text extract is a detailed discussion of a mathematical problem and its proof. It starts with a problem statement about proving the differentiability of an inverse function under certain conditions. The user mentions finding the proof in Spivak's Calculus book confusing and asks if there is an alternative proof. The extract then provides the proof from Spivak's book and includes a comment from another user clarifying a notation issue. The text also includes an alternative explanation of the proof strategy.
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distilabel_metadata.content: Alternative proof for differentiability of inverse function?
  
  Problem Let $f$ be a continuous one-one function defined on an interval and suppose that $f$ is differentiable at $f^{-1}(b)$ with the derivative $f'(f^{-1}(x)) \neq 0$. Prove that $f^{-1}(x)$ is also differentiable at $b$.
  
  I read the proof in Spivak Calculus book, but it's quite confusing to me. Then I looked up online, and I saw exact the same proof everywhere. So I wonder is there an alternative proof to this problem that I'm not aware of?
  
  Here is the proof from Spivak book
  
  Theorem 5 (Calculus by Spivak 4th edition, page 237-238)
  Proof. Let $b = f(a)$. Then $$\displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - f^{-1}(b)}{h} = \displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - a}{h}$$ Now every number $b + h$ in the domain of $f^{-1}$ can be written in the form $$b + h = f(a + k)$$ for a unique $k$. Then $$\displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - a}{h} = \displaystyle\lim_{h\to 0}\dfrac{f^{-1}(f(a + k)) - a}{f(a + k) - b} = \displaystyle\lim_{h\to 0} \dfrac{k}{f(a + k) - f(a)}$$ We have $$f^{-1}(b + h) = a + k \Leftrightarrow k = f^{-1}(b + h) - f^{-1}(b)$$ Since $f$ is continuous and one-one, $f^{-1}$ is also continuous at $b$. This means that $k$ approaches $0$ as $h$ approaches $0$. Since $$\displaystyle\lim_{k\to 0} \dfrac{f(a + k) - f(a)}{k} = f'(a) = f'(f^{-1}(b)) \neq 0$$ this implies that $$(f^{-1})'(b) = \dfrac{1}{f'(f^{-1}(b))}$$
  
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  What do you mean by "$f(x)$ is differentiable"? Is that supposed to be "$f$ is differentiable at $f^{-1}(x)$"? – wj32 Dec 5 '12 at 8:41
  @wj32: Yes, I was a little sloppy since I thought it was very known proof. Let me fix that. – Chan Dec 5 '12 at 8:43
  I delete my comments since you edited. – Willie Wong Dec 5 '12 at 9:34
  
  The strategy of the proof is simple, and the difficulty lies in the existence of the limit for the incremental ration of $f^{-1}$. Indeed, if $f^{-1}$ is differentiable, then $f^{-1} \circ f = \operatorname{Id}$ implies $$Df^{-1}(f(x))Df(x)=1$$ for every $x$ in the domain of $f$, and you conclude.
  On the other hand, the existence of the derivative of $f^{-1}$ is a standard application of the change-of-variable rule for limits, which is in turn a corollary of the theorem on limits of composite functions.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a detailed discussion of a mathematical problem and its proof. It starts with a problem statement about proving the differentiability of an inverse function under certain conditions. The user mentions finding the proof in Spivak's Calculus book confusing and asks if there is an alternative proof. The extract then provides the proof from Spivak's book and includes a comment from another user clarifying a notation issue. The text also includes an alternative explanation of the proof strategy.
  
  Label: question-and-answer