Inverse Prime?

Let:

$z = f^{-1}(x)$ $\implies f(z) =x$ $\implies \mathrm{e}^{z}\ln{z} =x$

Differentiating both sides with respect to $x$ :

$\mathrm{e}^{z}\left(\frac{1}{z} + \ln{z}\right)\frac{dz}{dx} = 1$ $\implies \frac{dz}{dx} = \frac{1}{\mathrm{e}^{z}\left(\frac{1}{z} + \ln{z}\right)}$

Recall that

$z = f^{-1}(x)$ $\implies \boxed{\frac{d}{dx} \left( f^{-1}(x)\right) = \frac{1}{\mathrm{e}^{f^{-1}(x)}\left(\frac{1}{f^{-1}(x)} + \ln\left(f^{-1}(x)\right)\right)}}$

Let $y=f(x)$ . Then we have $y = e^x \ln x$ . $\implies \dfrac {dy}{dx} = e^x \ln x + \dfrac {e^x}x$ and $\dfrac {dx}{dy} = \dfrac 1{e^x \ln x + \frac {e^x}x}$ . Replace $y$ with x) and \(x with $f^{-1}(x)$ . we have:

$\frac {d f^{-1}(x)}{dx} = \boxed{\frac 1{e^{f^{-1}(x)}\left(\ln (f^{-1}(x)) + \frac 1{f^{-1}(x)}\right)}}$