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Question Number 181541 by mr W last updated on 26/Nov/22

$$solveforf\left(x\right)suchthat$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={\mathrm{f}}^{-1}\left(\mathrm{x}\right)$$

Commented by a.lgnaoui last updated on 26/Nov/22

$${\left[f^{-1}\left(x\right)\right]}^{{}^{\prime }}=\frac{1}{f^{{}^{\prime }}\left(x\right)f\left(x\right)}{f}^{\prime }\left(x\right)=f^{-1}\left(x\right)$$$${\left[f^{-1}\left(x\right)\right]}^{{}^{\prime }}=\frac{1}{f^{-1}\left(x\right)f\left(x\right)}$$$$f^{-1}\left(x\right){\left[f^{-1}\left(x\right)\right]}^{{}^{\prime }}=\frac{1}{f\left(x\right)}$$$$2\times {\left({\left[f^{-1}\left(x\right)\right]}^2\right)}^{{}^{\prime }}=\frac{1}{f\left(x\right)}\left(2\right)$$$$$$$$\Rightarrow ({\left[{\left(f^{{}^{\prime }}\left(x\right)\right]}^2\right)}^{{}^{\prime }}=\frac{1}{2f\left(x\right)}$$$$\text{Prime causes double exponent: use braces to clarify}$$$$.......$$

Commented by universe last updated on 26/Nov/22

$$solveforf\left(x\right)suchthat$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={\mathrm{f}}^{-1}\left(\mathrm{x}\right)$$

Answered by a.lgnaoui last updated on 26/Nov/22

$$f^{\prime }\left(x\right)\times f\left(x\right)=1$$$${\left[\frac{{\left(f\left(x\right)\right)}^2}{2}\right]}^{{}^{\prime }}=1\Rightarrow \frac{{\left[f\left(x\right)\right]}^2}{2}=x+C$$$$f\left(x\right)=\sqrt{2x+k}k\in \mathbb{R}$$

Commented by a.lgnaoui last updated on 26/Nov/22

$$ohyes$$

Commented by cortano1 last updated on 26/Nov/22

$${\mathrm{f}}^{-1}\left(\mathrm{x}\right)\ne \frac{1}{\mathrm{f}\left(\mathrm{x}\right)}$$

Answered by cortano1 last updated on 26/Nov/22

$$\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}={\mathrm{f}}^{-1}\left(\mathrm{x}\right)$$$$\mathrm{f}\left(\mathrm{x}\right)=\int {\mathrm{f}}^{-1}\left(\mathrm{x}\right)\mathrm{dx}$$$$[\mathrm{let}\mathrm{x}=\mathrm{f}\left(\mathrm{u}\right)\Rightarrow \mathrm{dx}=\mathrm{df}\left(\mathrm{u}\right)]$$$$\mathrm{f}\left(\mathrm{x}\right)=\int {\mathrm{f}}^{-1}\left(\mathrm{f}\left(\mathrm{u}\right)\right)\mathrm{df}\left(\mathrm{u}\right)$$$$\mathrm{f}\left(\mathrm{x}\right)=\int \mathrm{u}\mathrm{d}\left(\mathrm{f}\left(\mathrm{u}\right)\right)$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{u}\mathrm{f}\left(\mathrm{u}\right)-\int \mathrm{f}\left(\mathrm{u}\right)\mathrm{du}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}{\mathrm{f}}^{-1}\left(\mathrm{x}\right)-\mathrm{F}\left(\mathrm{f}\left(\mathrm{x}\right)\right)+\mathrm{c}$$

Commented by mr W last updated on 26/Nov/22

$$weshoulddeterminef\left(x\right).$$

Answered by universe last updated on 26/Nov/22

$$seequestion173624$$

Answered by mahdipoor last updated on 26/Nov/22

$$\frac{df}{dx}\times \frac{{df}^{-1}}{dx}=1\Rightarrow \frac{df}{dx}=f^{-1}=\frac{dx}{{df}^{-1}}\Rightarrow$$$$\Rightarrow f^{-1}{df}^{-1}=dx\Rightarrow \frac{1}{2}{\left(f^{-1}\left(x\right)\right)}^2=x+c$$$$f^{-1}\left(x\right)={\left[2(x+c)\right]}^{1/2}\Rightarrow x=f\left({\left[2(x+c)\right]}^{1/2}\right)$$

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