TITLE: Question about Inverse Functions and their derivatives
QUESTION [0 upvotes]: Let $f(x)=2x\ln(x)$. Compute $[f^{-1}]'(2e)$.

My work: I began by suggesting that the largest domain for which
$f(x)$ is one to one is $[1/e,∞]$. Tried to algebraically find
$f^{−1}(x)$ but didn't succeed.

REPLY [0 votes]: We 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
$$\left (\frac{1}{e}, -\frac{2}{e} \right ),$$
the inverse of $f$ will have two (real) branches, there being two branches when $-\dfrac{2}{e} \leqslant x < 0$ and one when $x \geqslant 0$. The inverse for $f$ therefore needs to be restricted: one for the domain $(0,1/e]$, the other for $[1/e,\infty)$. 
Method 1 - By inspection
As the two previous answers have noted, such a problem can be solved by inspection. If we denote $f^{-1}$ as the inverse of $f$ on the restricted domain of $[1/e,\infty)$, by observing that
$$f(e) = 2e \cdot \ln (e) = 2e,$$
then
$$f^{-1} (2e) = e$$. 
Since $f'(x) = 2 \ln x + 2$, from
$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))},$$
at the point $x = 2e$ we have
$$(f^{-1})'(2e) = \frac{1}{f'(f^{-1}(2e))} = \frac{1}{f'(e)} = \frac{1}{2 \ln (e) + 2} = \frac{1}{4}.$$
Method 2 - The Lambert W function way
An explicit inverse for your function $f$ can be found in terms of the Lambert W function.
To find the inverse of $f$ we set
$$x = 2y \ln y,$$
and solve for $y$. Doing so we have
\begin{align*}
\frac{x}{2y} &= \ln (y)\\
\exp \left (\frac{x}{2y} \right ) &= y\\
\frac{1}{y} \exp \left (\frac{x}{2y} \right ) &= 1\\
\frac{x}{2y} \exp \left (\frac{x}{2y} \right ) &= \frac{x}{2}.
\end{align*}
As this last expression is now exactly in the form for the definiting equation for the Lambert W function, namely
$$\text{W} (x) e^{\text{W}(x)} = x,$$
we can write
$$\frac{x}{2y} = \text{W}_\nu \left(\frac{x}{2} \right),$$
or
$$y = \frac{x}{2 \text{W}_\nu \left (\frac{x}{2} \right )}.$$
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 again making use of the defining equation for the Lambert W function this can be expressed as
$$y = \exp \left [\text{W}_\nu \left (\frac{x}{2} \right ) \right ].$$
So the inverse of $f$ is given by
$$f^{-1} (x) = \begin{cases}
\exp \left [\text{W}_0 \left (\dfrac{x}{2} \right ) \right ], & x \geqslant -\dfrac{2}{e}\\[2ex]
\exp \left [\text{W}_{-1} \left (\dfrac{x}{2} \right ) \right ], & -\dfrac{2}{e} \leqslant x < 0.
\end{cases}$$
Now finding its derivative, noting that
$$\frac{d}{dx} \text{W}(x) = \frac{1}{x + e^{\text{W} (x)}},$$
on differentiating we have
$$(f^{-1})' (x) = \frac{\exp \left [\text{W}_\nu \left (\dfrac{x}{2} \right ) \right ]}{x + 2\exp \left [\text{W}_\nu \left (\dfrac{x}{2} \right ) \right ]},$$
where $\nu = -1,0$. 
For $x = 2e$ the principal branch is selected and we have
$$(f^{-1})' (2e) = \frac{\exp [\text{W}_0 (e)]}{2e + \exp [\text{W}_0 (e)]}.$$
Since $\text{W}_0(e) = 1$, the expression for the derivative of the inverse function at $x = 2e$ reduces to
$$(f^{-1})'(2e) = \frac{e}{2e + 2e} = \frac{1}{4},$$
as expected.
Inverse derivative of a function
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