# Question about derivatives of inverses functions.

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 $g^{\prime}\left(3\right)$?

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

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

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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

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, $$f'(6) g'(3) = 1 \implies g'(3) = \dfrac1{f'(6)} = -\dfrac12$$

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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:

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

$$g(x) = f^{-1}(x) = \frac{1}{f(x)} \iff g(x) = \frac{1}{2x} \neq \frac{x}{2}.$$
And so your assumption that $\displaystyle g^{\prime}(x) = \frac{1}{f^{\prime}(x)}$ cannot possibly be correct, at least most of the time!