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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?

Hosein Rahnama
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luka5z
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  • 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.......... – Jasser Dec 17 '15 at 16:41
  • @Jasser What do you mean by that? – luka5z Dec 17 '15 at 16:42
  • 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 $f^{-1}$ – Jasser Dec 17 '15 at 16:46
  • Could you post it as an answer? – luka5z Dec 17 '15 at 16:47

3 Answers3

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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

HeroZhang001
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Jasser
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  • Seems good, I have never seen it proved in this particular way though. – luka5z Dec 17 '15 at 17:06
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    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 $k\to 0$. – Giuseppe Negro Dec 17 '15 at 17:07
  • 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 – Jasser Dec 17 '15 at 17:16
  • But I dont know how do we prove it mathematically @GiuseppeNegro I think you can help here!! – Jasser Dec 17 '15 at 17:22
  • 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. – Giuseppe Negro Dec 17 '15 at 17:32
  • How do we approach to prove it @GiuseppeNegro .... By the way I could be wrong but $f$ has to be monotonic because the inverse function will not exist otherwise.... – Jasser Dec 17 '15 at 17:41
  • A continuous and invertible function between intervals is automatically monotonic, but this another theorem. We already know that our function is monotonic so there is no need for this. – Giuseppe Negro Dec 17 '15 at 17:42
  • Haha! Yes moreover if $f$ is continuous in an interval, mapping of continuous interval to another countinuos interval(finite intervals) on real line can also give an inverse mapping(assuming it exists) which is continuous and I think this can be proved by applying the epsion-delta method reversely. @GiuseppeNegro – Jasser Dec 17 '15 at 17:50
  • Now $f^{-1}(x+h)$ is $y+k$ and $f^{-1}(x)$ is $y$ -- should this instead be "$f(x+h)$ is $y+k$ and $f(x)$ is $y$" – user182601 Feb 01 '24 at 05:50
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This theorem is difficult to prove, mainly due to justifying the variable substitution of limits when applying the limit definition of the derivative. We state and prove the following formulation -

Let $f$ be a real-valued function defined on an open interval $I$, such that $f$ is both continuous and injective. Suppose $f$ is differentiable on $I$ with $f'(x) = 0$ for all $x \in I$. Then, $f^{-1}$ is differentiable, with: $$(f^{-1})'(b) = \dfrac{1}{f'(a)},$$ where $b=f(a)$ for all $b \in \mathrm{range}(f)$. Note that this value of $a$ is unique in $I$.

Lemma $1$ : Let $f$ be a continuous and injective function on an interval $I$. Then, $f^{-1}$ is also continuous and injective on $I$.

Lemma $2$ : Suppose $g$ is continuous and injective on an open interval $I$ containing $a$, and $f$ is some function. Then, $\displaystyle \lim_{x \rightarrow a} (f \circ g)(x)$ exists $\iff \displaystyle \lim_{y \rightarrow g(a)} f(y)$ exists.

For a proof of this Lemma, see the excellent answer in this post https://math.stackexchange.com/a/3852080/646705. Essentially, this lemma imposes additional constraints on the inner function of the composition when continuity of the outer function cannot be assumed. This is important when trying to perform a variable substitution of limits on an expression such as the difference quotient, which we will soon see.

Proof of Theorem : \begin{align*} \dfrac{1}{f'(a)} & = \dfrac{1}{\displaystyle \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}} = \displaystyle \lim_{x \rightarrow a} \frac{x-a}{f(x)-f(a)} = \displaystyle \lim_{x \rightarrow a} \frac{f^{-1}(f(x))-a}{f(x)-f(a)}. \end{align*}

Let $h(y) = \dfrac{f^{-1}(y)-a}{y-f(a)}$. Since $f$ is by assumption continuous and injective on $I$, applying Lemma $2$, \begin{align*} \dfrac{1}{f'(a)} & = \displaystyle \lim_{x \rightarrow a} \ (h \circ f)(x) = \displaystyle \lim_{y \rightarrow f(a)} \ h(y) \\ & = \lim_{y \rightarrow f(a)} \dfrac{f^{-1}(y) - f^{-1}(b)}{y-f(f^{-1}(b))} = \lim_{y \rightarrow b} \dfrac{f^{-1}(y) - f^{-1}(b)}{y-b} \\ & = (f^{-1})'(b), \end{align*} as desired.

Hello
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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 $\frac{d}{dx}(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}.$

HeroZhang001
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Brad A.M.
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