Consider $f(x) = 2x + \ln{x}$, $x>0$ and $g=f^{-1}$. Find the tangent line to the graph of $g$ at the point $(2, g(2))$.

The answer is given by $g'(2)$, so I have to find $g$, which is the inverse of $f$.

$$ f(x) = 2x + \ln{x} = \ln{e^{2x}} + \ln{x} = \ln{e^{2x} x} $$

And I couldn't continue from there, because I don't know how to isolate $x$. After searching it seems to be linked to something called "Lambert W function". Is there a way to solve this problem without evaluating $g$ directly?



Using the fact that $$(\forall x>0) \;\; (g\circ f)(x)=x$$ we get by differentiation

$$g'(f(x))f'(x)=1=g'(f(x))(2+\frac 1x)$$ and with $x=1$,


  • $\begingroup$ Cool, I had tried to differentiate $f(f^{-1}(x))=x$, but it didn't helped. I did not think of differentiating $f^{-1}(f(x))=x$! Thanks a lot! $\endgroup$ – rodorgas May 26 at 0:40

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