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I want to prove the following result:

$$ {f^{-1}}'(x)=\frac{1}{f'({f^{-1}}(x))}$$

Is simple application of chain rule a valid proof of it?

i.e. $$f({f^{-1}}(x))=x \implies \frac{df({f^{-1}}(x))}{dx}=1$$ and hence the result. Or is this not a standard proof? Is there additional conditions necessary, expect of course that the function is bijective, or that the inverse exists.

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You also need to apply the chain rule. – anthus Feb 27 '13 at 13:03
up vote 1 down vote accepted

The question is: Why is $f^{-1}$ differentiable? In the theorem you probably learned in a lecture this isn't assumed. However if you assume that $f^{-1}$ and $f$ are differentiable at and all the expressions make sense $f^{-1}$ should be a function and $f'\neq 0$ (both at least locally).

But nevertheless, with the chainrule the formula is easy to remember.

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Let $y_0\in \mathbb{R} $ then there's $x_0\in\mathbb{R}$ such that $f(x_0)=y_0$. If $x\rightarrow x_0$ and by continuity of $f$ (because $f$ is differentiable) we have $y=f(x)\rightarrow f(x_0)=y_0$. Now, let's check the differentiability of $f^{-1}$ by the definition: $$(f^{-1})'(y_0)=\lim_{y\rightarrow y_0}\frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=\lim_{y\rightarrow y_0}\frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\\=\lim_{x\rightarrow x_0}\frac{x-x_0}{f(x)-f(x_0)}=\frac{1}{f'(x_0)}=\frac{1}{f'(f^{-1}(y_0))}.$$

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For differentiable functions $\mathbb{R} \to \mathbb{R}$ it is easy using geometrical interpretation. Take the graph of $f$, then the graph of $f^{-1}$ is symmetrical with respect to $\{y = x\}$ line. However, $f'(x) = \tan \alpha$ implies that $${f^{-1}}'(y) = \tan\left(\frac{\pi}{2}-\alpha\right) = \cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{f'(x)}.$$

Hope that helps ;-)

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The chain rule "derivation" is invalid, because it assumes $f{^-1}$ is differentiable, which is begging the question (circular reasoning).

Here is a valid proof:


i) Function $f$ defined and continuous on a neighborhood of point $x$.

ii) Inverse function $f{^-1}$ defined and continuous on a neighborhood of $y = f(x)$.

iii) $f$ differentiable at point $x$, and $f'(x) \not = 0$.

By the differentiability theorem:

$$f(x + h) - f(x) = h(f'(x) + g(h))$$

where $g(h)$ goes to zero as h goes to zero.

define $k := h(f'(x) + g(h))$

by limit theorem $k$ also goes to zero as $h$ does.

Since $f'(x)$ is non-zero, you can restrict $|h|$ so that that $|g(h)|$ is less than $|f'(x)|$ (and thus k is non-zero), and add a further restriction so that $f^-1$ is defined and continuous on the interval $[y-k, y + k]$.

Now you can show that the derivative exists:

$$\frac{1}{f'(x)} = \lim_{h\to 0} \frac{h}{f(x+h) - f(x)} = \lim_{h\to 0} \frac{f^{-1}(f(x + h)) - f^{-1}(f(x))}{f(x+h)-f(x)} = \lim_{k\to 0} \frac{f^{-1}(y+k) - f^{-1}(y)}{k}$$

Note: If a function is continuous and strictly increasing or strictly decreasing on a neighborhood of $x$, the inverse function theorem says that $f^{-1}$ is then also strictly increasing/decreasing on a neighborhood of $y$. Thus the inverse function is defined and continuous on a neighborhood of $y$.

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