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Question Number 150594 by liberty last updated on 13/Aug/21

The function f (x) = e^{x} + x being

differentiable and one to one,

has a differentiable inverse f^{- 1} (x) .

The value of \frac{d}{d x} (f^{- 1}) at point

f (\ln 2) is__

Answered by Olaf_Thorendsen last updated on 13/Aug/21

f (x) = e^{x} + x

f^{'} (x) = e^{x} + 1

f o f^{- 1} (x) = x

{(f^{- 1})}^{'} (x) \times f^{'} o f^{- 1} (x) = 1

{(f^{- 1})}^{'} (x) = \frac{1}{f^{'} o f^{- 1} (x)}

{(f^{- 1})}^{'} (f (\ln 2)) = \frac{1}{f^{'} o f^{- 1} (f (\ln 2))}

{(f^{- 1})}^{'} (f (\ln 2)) = \frac{1}{f^{'} (\ln 2)} = \frac{1}{e^{\ln 2} + 1} = \frac{1}{3}