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Question-202761

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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(x)=f−1(x)f(x)=g−1(x)f′(x)=sin(x)f(x)=c−cos(x)wherecisconstant.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)=1sin(g(x))=1sin(cos−1(c−x))wherecisconstant.★

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