Question Number 181541 by mr W last updated on 26/Nov/22
$${solve}\:{for}\:{f}\left({x}\right)\:{such}\:{that} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$
Commented by a.lgnaoui last updated on 26/Nov/22
![[f^(−1) (x)]^′ =(1/(f^′ (x)f(x))) f′(x)=f^(−1) (x) [f^(−1) (x)]^′ =(1/(f^(−1) (x)f(x))) f^(−1) (x)[f^(−1) (x)]^′ =(1/(f(x))) 2×([f^(−1) (x)]^2 )^′ =(1/(f(x))) (2) ⇒([(f^′ (x)]^2 )^′ =(1/(2f(x))) 2f^′ ′(x)×[f^′ (x)]^2 =(1/(2f(x))) .......](https://www.tinkutara.com/question/Q181549.png)
$$\left[{f}^{−\mathrm{1}} \left({x}\right)\right]^{'} =\frac{\mathrm{1}}{{f}^{'} \left({x}\right){f}\left({x}\right)}\:\:\:\:\:\:\:{f}'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left[{f}^{−\mathrm{1}} \left({x}\right)\right]^{'} =\frac{\mathrm{1}}{{f}^{−\mathrm{1}} \left({x}\right){f}\left({x}\right)} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\left[{f}^{−\mathrm{1}} \left({x}\right)\right]^{'} =\frac{\mathrm{1}}{{f}\left({x}\right)} \\ $$$$\mathrm{2}×\left(\left[{f}^{−\mathrm{1}} \left({x}\right)\right]^{\mathrm{2}} \right)^{'} =\frac{\mathrm{1}}{{f}\left({x}\right)}\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\Rightarrow\left(\left[\left({f}^{'} \left({x}\right)\right]^{\mathrm{2}} \right)^{'} =\frac{\mathrm{1}}{\mathrm{2}{f}\left({x}\right)}\right. \\ $$$$\mathrm{2}{f}^{'} '\left({x}\right)×\left[{f}^{'} \left({x}\right)\right]^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}{f}\left({x}\right)} \\ $$$$……. \\ $$
Commented by universe last updated on 26/Nov/22
$${solve}\:{for}\:{f}\left({x}\right)\:{such}\:{that} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$
Answered by a.lgnaoui last updated on 26/Nov/22
![f′(x)×f(x)=1 [(((f(x))^2 )/2)]^′ =1⇒ (([f(x)]^2 )/2)=x+C f(x)=(√(2x+k)) k∈R](https://www.tinkutara.com/question/Q181543.png)
$${f}'\left({x}\right)×{f}\left({x}\right)=\mathrm{1} \\ $$$$\left[\frac{\left({f}\left({x}\right)\right)^{\mathrm{2}} }{\mathrm{2}}\right]^{'} =\mathrm{1}\Rightarrow\:\frac{\left[{f}\left({x}\right)\right]^{\mathrm{2}} }{\mathrm{2}}={x}+{C} \\ $$$${f}\left({x}\right)=\sqrt{\mathrm{2}{x}+{k}}\:\:\:\:\:{k}\in\mathbb{R}\: \\ $$
Commented by a.lgnaoui last updated on 26/Nov/22
$${oh}\:{yes} \\ $$
Commented by cortano1 last updated on 26/Nov/22
$$\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\:\neq\:\frac{\mathrm{1}}{\mathrm{f}\left(\mathrm{x}\right)} \\ $$
Answered by cortano1 last updated on 26/Nov/22
![((df(x))/dx) = f^(−1) (x) f(x)=∫ f^(−1) (x) dx [ let x = f(u) ⇒dx=df(u) ] f(x)= ∫ f^(−1) (f(u)) df(u) f(x)= ∫ u d(f(u)) f(x)= u f(u)−∫ f(u)du f(x)= x f^(−1) (x)−F(f(x)) + c](https://www.tinkutara.com/question/Q181542.png)
$$\:\:\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$$$\:\:\mathrm{f}\left(\mathrm{x}\right)=\int\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\:\mathrm{dx} \\ $$$$\:\left[\:\mathrm{let}\:\mathrm{x}\:=\:\mathrm{f}\left(\mathrm{u}\right)\:\Rightarrow\mathrm{dx}=\mathrm{df}\left(\mathrm{u}\right)\:\right] \\ $$$$\:\:\mathrm{f}\left(\mathrm{x}\right)=\:\int\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{f}\left(\mathrm{u}\right)\right)\:\mathrm{df}\left(\mathrm{u}\right)\: \\ $$$$\:\:\mathrm{f}\left(\mathrm{x}\right)=\:\int\:\mathrm{u}\:\mathrm{d}\left(\mathrm{f}\left(\mathrm{u}\right)\right) \\ $$$$\:\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{u}\:\mathrm{f}\left(\mathrm{u}\right)−\int\:\mathrm{f}\left(\mathrm{u}\right)\mathrm{du} \\ $$$$\:\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{x}\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)−\mathrm{F}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\:+\:\mathrm{c}\: \\ $$
Commented by mr W last updated on 26/Nov/22
$${we}\:{should}\:{determine}\:{f}\left({x}\right). \\ $$
Answered by universe last updated on 26/Nov/22
$${see}\:{question}\:\mathrm{173624} \\ $$
Answered by mahdipoor last updated on 26/Nov/22
![(df/dx) × (df^(−1) /dx) =1 ⇒ (df/dx)=f^(−1) =(dx/df^(−1) ) ⇒ ⇒f^(−1) df^(−1) =dx ⇒(1/2)(f^(−1) (x))^2 =x+c f^(−1) (x)=[2(x+c)]^(1/2) ⇒ x=f([2(x+c)]^(1/2) )](https://www.tinkutara.com/question/Q181560.png)
$$\frac{{df}}{{dx}}\:×\:\frac{{df}^{−\mathrm{1}} }{{dx}}\:=\mathrm{1}\:\:\Rightarrow\:\:\frac{{df}}{{dx}}={f}^{−\mathrm{1}} =\frac{{dx}}{{df}^{−\mathrm{1}} }\:\Rightarrow \\ $$$$\Rightarrow{f}^{−\mathrm{1}} {df}^{−\mathrm{1}} ={dx}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left({f}^{−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} ={x}+{c} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\left[\mathrm{2}\left({x}+{c}\right)\right]^{\mathrm{1}/\mathrm{2}} \:\Rightarrow\:{x}={f}\left(\left[\mathrm{2}\left({x}+{c}\right)\right]^{\mathrm{1}/\mathrm{2}} \right) \\ $$