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

$$\mathrm{The}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}+\mathrm{x}\mathrm{being}$$$$\mathrm{differentiable}\mathrm{and}\mathrm{one}\mathrm{to}\mathrm{one},$$$$\mathrm{has}\mathrm{a}\mathrm{differentiable}\mathrm{inverse}{\mathrm{f}}^{-1}\left(\mathrm{x}\right).$$$$\mathrm{The}\mathrm{value}\mathrm{of}\frac{d}{dx}\left(f^{-1}\right)\mathrm{at}\mathrm{point}$$$$\mathrm{f}\left(\ln 2\right)\mathrm{is}\mathrm{\_}\mathrm{\_}$$

Answered by Olaf_Thorendsen last updated on 13/Aug/21

$$f\left(x\right)=e^x+x$$$$f^{\prime }\left(x\right)=e^x+1$$$$f\mathrm{o}{f}^{-1}\left(x\right)=x$$$${\left(f^{-1}\right)}^{\prime }\left(x\right)\times f^{\prime }\mathrm{o}{f}^{-1}\left(x\right)=1$$$${\left(f^{-1}\right)}^{\prime }\left(x\right)=\frac{1}{f^{\prime }\mathrm{o}{f}^{-1}\left(x\right)}$$$${\left(f^{-1}\right)}^{\prime }\left(f\left(\ln 2\right)\right)=\frac{1}{f^{\prime }\mathrm{o}{f}^{-1}\left(f\left(\ln 2\right)\right)}$$$${\left(f^{-1}\right)}^{\prime }\left(f\left(\ln 2\right)\right)=\frac{1}{f^{\prime }\left(\ln 2\right)}=\frac{1}{e^{\ln 2}+1}=\frac{1}{3}$$

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