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.
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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.
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The text passage contains a mathematical problem statement and a detailed proof of a theorem from calculus, but it does not contain a mathematical question in the form of a word problem or calculation request, nor does it provide an answer to such a question.

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