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?

  • $\begingroup$ 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


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

  • $\begingroup$ Seems good, I have never seen it proved in this particular way though. $\endgroup$
    – luka5z
    Dec 17, 2015 at 17:06
  • 2
    $\begingroup$ 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

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

  • 4
    $\begingroup$ But you have to know that $f^{-1}$ is differentiable $\endgroup$
    – luka5z
    Dec 18, 2015 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.