Inverse derivative of a function

Question

Problem

We have function defined as:

$$ f(x)=x^2\ln(x)$$

What is

$$ (f^{-1})'(y_0) $$ on specific point which is $y_0=e^2$

Attempt to solve

A derivative of a inverse function would be defined as:

$$ (f^-1)'(y_0)=\frac{1}{f'(f^{-1}(y_0))} $$

We can easily define $f'(x)$ $$ f'(x)=x+2x\ln(x) $$

now the problem is we don't know what is $f^{-1}$. It seems it's not possible to compute derivative of inverse function without first computing the inverse of the function. Now i don't know to to solve inverse function of $f(x)=x^2\ln(x)$ but i can see for example what wolframalpha would suggest.

wolframalpha suggests that inverse function would be something like this:

$$f^{-1}=\pm \frac{\sqrt{2}\sqrt{x}}{\sqrt{W(2x)}}$$

where W is Lambert W function

Now combining these solutions.

$$ (f^{-1})'(y_0)=\frac{1}{(\pm \frac{\sqrt{2}\sqrt{y_0}}{\sqrt{W(2y_0)}})+2 (\pm \frac{\sqrt{2}\sqrt{y_0}}{\sqrt{W(2y_0)}})\ln(\pm \frac{\sqrt{2}\sqrt{y_0}}{\sqrt{W(2y_0)}})} $$

Now the problem is i don't know how to deal with the Lambert-W function or so called "omega function". It seems i need some help solving this problem.

It would be highly appreciated if someone could hint me in right direction. At this point i am pretty convinced this is not the best way to solve this problem.

Answer 1

You are looking for

$$f^{-1}(e^2) = k$$ By definition $$f(k) = e^2 = k^2\ln k$$

And this is quite easy to see that $k=e$

From here,

$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$$ $$(f^{-1})'(e^2) = \frac{1}{f'(e)} = \frac{1}{e + 2e\ln e} = \frac{1}{3e}$$

Answer 2

As @John Lou points out, these problems, by design, are expected to be done by inspection - the point having been carefully selected so that everything works out.

But what would happen if this was not so? For example, how would one find the value of $(f^{-1})' (x)$ at say $x = 1/2$?

As you correctly suggest, the inverse function in terms of the Lambert W function would need to be found.

First note that as $x \to 0^+$, $f(x) \to 0$, as $x \to \infty$, $f(x) \to \infty$, and since $f$ has a single (global) minimum at $$\left (\frac{1}{\sqrt{e}}, -\frac{1}{2e} \right )$$ the inverse of $f$ will have two (real) branches, there being two branches when $-\frac{1}{2e} \leqslant x < 0$ and one when $x \geqslant 0$.

To find the inverse of $f$ we set $$x = y^2 \ln y,$$ and solve for $y$. Doing so leads to $$y^2 = \frac{2x}{\text{W}_\nu (2x)}.$$ Here $\nu$ denotes the two real branches ($\nu = 0$ corresponding to the principal branch; $\nu = -1$ corresponding to the secondary real branch) for the Lambert W function. By making use of the defining equation for the Lambert W function, namely $$\text{W} (x) e^{\text{W}(x)} = x,$$ we can write $$\frac{2x}{\text{W} (2x)} = e^{\text{W}(2x)},$$ so that $$y^2 = \exp \left (\text{W}(2x) \right ).$$

Thus $$f^{-1} (x) = \begin{cases} \exp \left (\frac{1}{2} \text{W}_0 (2x) \right ), & x \geqslant -\dfrac{1}{2e}\\[2ex] \exp \left (\frac{1}{2} \text{W}_{-1} (2x) \right ), & -\dfrac{1}{2e} \leqslant x < 0. \end{cases}$$

Now finding its derivative, noting that $$\frac{d}{dx} \text{W}(x) = \frac{1}{x + e^{\text{W} (x)}},$$ on differentiating we have $$(f^{-1})' (x) = \frac{\exp \left [\frac{1}{2} \text{W}_\nu (2x) \right ]}{2x + \exp \left [\text{W}_\nu (2x) \right ]},$$ where $\nu = -1,0$.

For $x = e^2$ the principal branch is selected and we have $$(f^{-1})' (e^2) = \frac{\exp \left [\frac{1}{2} \text{W}_0 (2e^2) \right ]}{2e^2 + \exp \left [\text{W}_0 (2e^2) \right ]}.$$

As $\text{W} (x)$ is the inverse of the function $x e^x$ we have the following simplification rule for the Lambert W function of $$\text{W}_0 (x e^x) = x, \quad x \geqslant -1.$$ Thus $\text{W}_0 (2e^2) = 2,$ and we have $$(f^{-1})' (e^2) = \frac{e}{2e^2 + e^2} = \frac{1}{3e},$$ as expected.

A far more interesting exercise however is to now find the value of the derivative of the inverse function at $x = 1/2$, which, as I pointed out at the beginning, is an example that can no longer be done using inspection.

Here we have $$(f^{-1})' \left (\frac{1}{2} \right ) = \frac{\exp [\text{W}_0 (1)/2]}{1 + \exp [\text{W}_0 (1)]}.$$ The value of the Lambert W function when its argument is equal to one corresponds to the so-called omega constant, that is, $\text{W}_0 (1) = \Omega$. Thus $$(f^{-1})' \left (\frac{1}{2} \right ) = \frac{e^{\Omega/2}}{1 + e^\Omega} = \frac{1}{e^{\Omega/2} + e^{-\Omega/2}} = \frac{1}{2 \cosh \left (\frac{\Omega}{2} \right )} = \frac{1}{2} \text{sech} \left (\frac{\Omega}{2} \right ).$$