Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let be a differentiable function such that $f\left(3\right)=15$, $f\left(6\right)=3$ ,$f^{\prime}\left(6\right) = -2$ , and $f^{\prime}\left(3\right) = -8$. The function $g\left(x\right)$ is differentiable and $g\left(x\right) = f^{-1}\left(x\right)$ for all $x$. What is the value of $g^{\prime}\left(3\right)$?

My answer:

Because $g^{\prime}\left(x\right) = \frac{1}{f^{\prime}\left(x\right)}$

Thus $g^{\prime}\left(3\right) = \frac{1}{-8}$

I have realized my mistake.

I incorrectly understood the derivative of an inverse function. I should have stated

$$ \frac{d}{dx} f^{-1} = \frac{1}{f^{\prime}\left(f^{-1}\left(x\right)\right)} $$

share|cite|improve this question
I think your understanding about the inverse of a function is mistaken. One cannot assume that if $g(x) = f^{-1}(x)$, that $g(x) = \frac{1}{f(x)}$. That works for constant functions (except for $f(x) = 0$), but is seldom true otherwise. It then seems that, based on your mistaken understanding of an inverse of a function, you then assumed it must be the case that $g^{\prime}(x) = \frac{1}{f^{\prime}(x)}$ – amWhy Nov 7 '12 at 0:46

2 Answers 2

up vote 1 down vote accepted

We have $$f(g(x)) = x$$ Hence, by chain rule, we get that $$\dfrac{df(g(x))}{dg(x)} \dfrac{dg(x)}{dx} = 1$$ When $x=3$, we have that $g(3) = f^{-1}(3) = 6$. Hence, we get that $$\left. \dfrac{df(g(x))}{dg(x)} \cdot \dfrac{dg(x)}{dx} \right \vert_{x=3} = \left. \dfrac{df(y)}{dy} \right \vert_{y=g(3) = 6} \left. \dfrac{dg(x)}{dx} \right \vert_{x=3} = 1$$ Hence, $$f'(6) g'(3) = 1 \implies g'(3) = \dfrac1{f'(6)} = -\dfrac12$$

share|cite|improve this answer

To elaborate on my comment regarding your (mistaken) understanding of what the inverse of a function means, let me clarify using a counterexample.

Suppose $f(x) = 2x\;\; \text{and}\;\;g(x) = f^{-1}(x) = \frac{x}{2}$.

You can confirm that $g(x)$ is indeed the inverse of $f(x)$ by checking:

$$f(g(x)) = f\left(\frac{x}{2}\right) = 2\left(\frac{x}{2}\right) = x.$$

Can you see that your faulty reasoning leads to the following contradiction?:

$$g(x) = f^{-1}(x) = \frac{1}{f(x)} \iff g(x) = \frac{1}{2x} \neq \frac{x}{2}.$$

And so your assumption that $\displaystyle g^{\prime}(x) = \frac{1}{f^{\prime}(x)}$ cannot possibly be correct, at least most of the time!

Perhaps you should clarify your understanding of a function and its inverse. Then understanding how those functions' derivatives relate will make more sense to you.

share|cite|improve this answer
I just installed now Mathematica 9 on my Ubuntu and it seems powerful. I saw a quick tour of it on youtube and it's really fantastic! Do you suggest any book that can be a good introduction for me to Mathematica? – user63181 Jul 16 '14 at 11:22
@Sami Mathematica is powerful. I just got 10, (got 9 last year and was able to upgrade), But I still need to put my head to the task of learning how to use it! It is powerful, and I believe Wolfram has a site with tutorials, etc. Maybe we can join forces on our missions to conquer using it? ;-) – amWhy Jul 16 '14 at 11:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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