Question Number 202761 by Mastermind last updated on 02/Jan/24
Answered by shunmisaki007 last updated on 03/Jan/24
$${g}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right) \\ $$$${f}\left({x}\right)={g}^{−\mathrm{1}} \left({x}\right) \\ $$$${f}'\left({x}\right)=\mathrm{sin}\left({x}\right) \\ $$$${f}\left({x}\right)={c}−\mathrm{cos}\left({x}\right)\:\mathrm{where}\:{c}\:\mathrm{is}\:\mathrm{constant}. \\ $$$${f}\left({g}\left({x}\right)\right)={c}−\mathrm{cos}\left({g}\left({x}\right)\right)={g}^{−\mathrm{1}} \left({g}\left({x}\right)\right) \\ $$$$\mathrm{cos}\left({g}\left({x}\right)\right)={c}−{x}\:\left(\Rightarrow{g}\left({x}\right)=\mathrm{cos}^{−\mathrm{1}} \left({c}−{x}\right)\right) \\ $$$$−\mathrm{sin}\left({g}\left({x}\right)\right){g}'\left({x}\right)=−\mathrm{1} \\ $$$$\therefore\:{g}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{sin}\left({g}\left({x}\right)\right)}=\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{cos}^{−\mathrm{1}} \left({c}−{x}\right)\right)}\:\mathrm{where}\:{c}\:\mathrm{is}\:\mathrm{constant}.\:\bigstar \\ $$$$ \\ $$