# How can I find the derivative of the inverse of f(x)= x^3+x+1 at x=11?

Jun 17, 2015

The answer is $\frac{1}{13}$.

#### Explanation:

For an initial explanation, I think it's better not to"swap" the variables in writing the inverse function. In other words, if $x$ is the independent (input) variable for $f$, it will become the dependent (output) variable for ${f}^{- 1}$. (This is important in applied problems so that the variable names have consistent meanings).

Let $y = f \left(x\right) = {x}^{3} + x + 1$ and let $x = {f}^{- 1} \left(y\right)$ be the inverse function of $f$ (there's no need to find a formula for ${f}^{- 1}$).

The rule for differentiating the inverse function in this context can be written in Leibniz notation as $\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}}$. While this is somewhat intuitive, it is a bit sloppy because it's not clear how to use it. To be more precise, we can write the rule as (f^{-1})'(y)=1/(f'(f^{-1}(y)).

Note that $f ' \left(x\right) = 3 {x}^{2} + 1$ and that $f \left(2\right) = {2}^{3} + 2 + 1 = 11$ so that ${f}^{- 1} \left(11\right) = 2$. Then the rule at the end of the paragraph above can be written as $\left({f}^{- 1}\right) ' \left(11\right) = \frac{1}{f ' \left({f}^{- 1} \left(11\right)\right)} = \frac{1}{f ' \left(2\right)} = \frac{1}{3 \cdot {2}^{2} + 1} = \frac{1}{13}$. That's the answer. Notice that I took the number $11$ to be a $y$-value instead of an $x$-value because I did not "swap" the $x$ and $y$ in writing the inverse function.

If you don't do the swapping of the variables, here's an intuitive justification of what I've done. Letting $y = f \left(x\right)$ and noting that $f \left(2\right) = 11$ and $f ' \left(2\right) = 13$, we get the approximation $\setminus \Delta y \setminus \approx 13 \cdot \setminus \Delta x$ when $x = 2$ and $\setminus \Delta x \setminus \approx 0$. This means, for instance, that $f \left(2.01\right) \setminus \approx 11 + 13 \cdot 0.01 = 11.13$.

Since $\Delta y \setminus \approx 13 \cdot \setminus \Delta x$ in such a situation, it should make intuitive sense that $\Delta x \setminus \approx \frac{1}{13} \cdot \setminus \Delta y$ in essentially the same situation (thought of this time in terms of $y = 11$ and $\setminus \Delta y \setminus \approx 0$). This means, for example, that ${f}^{- 1} \left(11.01\right) \setminus \approx 2 + \frac{1}{13} \cdot 0.01 \setminus \approx 2.00077$.

If you do swap the variables and write $y = f \left(x\right)$ as well as $y = {f}^{- 1} \left(x\right)$, then the well-known reflection property of the graphs of these functions across the line $y = x$ is what gives geometric justification of what I've done.

Here's the picture in the given situation. In this picture the graph of $y = f \left(x\right) = {x}^{3} + x + 1$ is shown in blue, the point $\left(x , y\right) = \left(2 , 11\right)$ on it is shown, as well as the tangent line with a slope of $13$. The graph of $y = {f}^{- 1} \left(x\right)$ is shown in green, the point $\left(x , y\right) = \left(11 , 2\right)$ on it is shown, as well as the tangent line with a slope of $\frac{1}{13}$. These tangent lines have slopes that are reciprocals of each other because of the reflection property across the line $y = x$.