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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.

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  • $\begingroup$ 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. $\endgroup$ – Arth Gupta Dec 8 '17 at 8:46
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Using the formula for derivative of inverse function we have:

$$ [f^{-1}]'(2e) = \frac{1}{f'(f^{-1}(2e))} $$ $f'$ shouldn't be hard to compute. It remains to find $f^{-1}(2e)$ which is equivalent to solve the equation $2x\ln(x) = 2e$. This is not a trivial equation indeed, but by simply guessing, you should easily find the solution.

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You need not to find $f^{-1}(x)$ for any $x$, but just $f^{-1}(2e)$. Recall that (see for example Inverse functions and differentiation), $$[f^{-1}]'(2e)=\frac{1}{f'(f^{-1}(2e))},$$ where $f^{-1}$ is the inverse of $f$ restricted to the domain $[1/e,+\infty)$. Can you take it from here?

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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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