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Question-202761




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

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