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Let $I$ be an open interval in $\mathbb{R}$ that contains point $a$ and let $f:I\rightarrow \mathbb{R}$ be a continuously differentiable function such that $f'(a)\ne 0$. Then there exist open intervals $U\subset I$ and $V$ in $\mathbb{R}$ such that restriction of $f$ to $U$ is a bijection of $U$ onto $V$ whose inverse $f^{-1}: V\rightarrow U$ is differentiable.

Since $f'(a)$ is non zero $f'$ is continuous, it must be the case that $f'$ does not change sign in some small ball $U=(a-\epsilon, a+\epsilon)\subset I$. Thus $f$ is strictly increasing (decreasing) on that ball. Therefore $f$ restricted to $U_0$ is a bijection from $U_0$ onto $V=f(U)$. Thus there is $f^{-1}:V\rightarrow U$ which is a continuous bijection. We can use mean value theorem, so for $x,x_1 \in V$ such that $x\ne x_1$ there is some $\theta$ between $f^{-1}(x)$ and $f^{-1}(x_1)$ such that

$$x-x_1=f(f^{-1}(x))-f(f^{-1}(x_1))=(f^{-1}(x)-f^{-1}(x_1)) f'(\theta)$$

thus $$\frac{f^{-1}(x)-f^{-1}(x_1)}{x-x_1}=\frac{1}{f'(\theta)}$$

(we know that $f'(\theta)$ is not zero)

And here is where I'm stuck. Any help please?

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  • $\begingroup$ I am not sure, but do we require mean value theorem here..... you can apply the first derivative principle to the inverse function and jump to the conclusion.......... $\endgroup$
    – Jasser
    Dec 17, 2015 at 16:41
  • $\begingroup$ @Jasser What do you mean by that? $\endgroup$
    – luka5z
    Dec 17, 2015 at 16:42
  • $\begingroup$ I meant to apply the definition of derivative to the inverse function and take the numerator to the denominator)....Also $x=f(y)$ where y belongs to the domain of f and x belongs to the domain of $f^{-1}$ $\endgroup$
    – Jasser
    Dec 17, 2015 at 16:46
  • $\begingroup$ Could you post it as an answer? $\endgroup$
    – luka5z
    Dec 17, 2015 at 16:47

2 Answers 2

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Let x and x+h belong to the domain of f where inverse exists and $f(x)=y$ and $f(x+h)=y+k$

Also $f^{-1}(y+k)=x+h$ and $f^{-1}(y)=x$

Limit as h tends to zero $\frac {f^{-1}(y+k)-f^{-1}(y)}{k}$

Now $f^{-1}(x+h)$ is $y+k$ and $f^{-1}(x)$ is $y$

so the limit becomes

$= \frac {h}{f(x+h)-f(x)}$ and now take h to the denominator and apply limit

Sorry I am bad at typing do it is difficult to write an answer

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  • $\begingroup$ Seems good, I have never seen it proved in this particular way though. $\endgroup$
    – luka5z
    Dec 17, 2015 at 17:06
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    $\begingroup$ This is correct but I would stress more clearly that we are using the (already proved) fact that $f^{-1}$ is continuous at $y$. Otherwise the change of variable $x+h=f^{-1}(y+k)$ might not imply that $h\to 0$ as $k\to 0$. $\endgroup$ Dec 17, 2015 at 17:07
  • $\begingroup$ When you find an inverse of a continuous function then it is also continuous.... To say intuitively we take the reflection of $f$ w.r.t the line $y=x$ which would imply $f^{=1}$ is also continuous @GiuseppeNegro $\endgroup$
    – Jasser
    Dec 17, 2015 at 17:16
  • $\begingroup$ But I dont know how do we prove it mathematically @GiuseppeNegro I think you can help here!! $\endgroup$
    – Jasser
    Dec 17, 2015 at 17:22
  • $\begingroup$ The inverse of a strictly monotone and continuous function $f\colon I\to J$, where $I, J$ are intervals, is indeed continuous but this is not an obvious statement and must be proved. (Actually one does not even need to assume that $f$ be strictly monotone). It is, however, something standard that can be found on (almost) all textbooks. $\endgroup$ Dec 17, 2015 at 17:32
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The chain rule can be used to show the derivative of the inverse is the inverse of the derivative. By composition of inverse functions we have $f(f^{-1}(x)=x.$ Differentiating both sides of the equation and applying the chain rule to the left hand side yields $f'(f^{-1}(x))\cdot \frac{d}{dx}(f^{-1}(x))=1.$ Then dividing gives $\frac{d}{dx}(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}.$

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    $\begingroup$ But you have to know that $f^{-1}$ is differentiable $\endgroup$
    – luka5z
    Dec 18, 2015 at 0:27

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