The function $f(x)=e^x+x$ being differentiable and one to one, has a differentiable inverse $f^{-1}(x)$, then find $\dfrac {d}{dx} (f^{-1}(x))$ at the point $f(log_e 2)$.

Let $y=e^x+x$

on differentiating w.r.t y,

$1=(e^x+1)\dfrac {dx}{dy}\Rightarrow \dfrac {dx}{dy}=\dfrac {1}{1+e^x}$

$\therefore \left (\dfrac {dx}{dy}\right )_{x=log_e2}=\dfrac {1}{1+e^{log 2}}=\dfrac {1}{3}$

or $\left [\dfrac {d}{dx}(f^{-1}(x))\right ]_{x=log 2}=\dfrac {1}{3}$