Given f(x)=ln(x)+exf(x) = \ln(x) + e^xf(x)=ln(x)+ex, find (f−1)′(1)(f^{-1})'(1)(f−1)′(1) where f(1)=ef(1) = ef(1)=e.
e/(e+1)e/(e+1)e/(e+1)
(e+1)/e(e+1)/e(e+1)/e
1/(e+1)1/(e+1)1/(e+1)
1/e1/e1/e