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Question Number 28858 by amit96 last updated on 31/Jan/18
Commented by abdo imad last updated on 31/Jan/18
$${we}\:{have}\:{f}\left({x}\right)={x}^{\mathrm{3}} +{x}\:{and}\:{g}\left({x}\right)={x}^{\mathrm{3}} −{x}\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\:{and}\:{g}^{'} \left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\:{and} \\ $$$$\left({gof}^{−\mathrm{1}} \right)^{'} \left(\mathrm{2}\right)={g}^{'} \left({f}^{−\mathrm{1}} \left(\mathrm{2}\right)\right).{f}^{−\mathrm{1}^{'} } \left(\mathrm{2}\right)\:{but}\:{f}^{−\mathrm{1}} \left(\mathrm{2}\right)={t}\:\Leftrightarrow{f}\left({t}\right)=\mathrm{2} \\ $$$$\Leftrightarrow{t}^{\mathrm{3}} +{t}\:−\mathrm{2}=\mathrm{0}\:\:\Leftrightarrow\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{2}\right)=\mathrm{0}\:\Leftrightarrow{t}=\mathrm{1}\:{because} \\ $$$${the}\:{polynomial}\:{t}^{\mathrm{2}} +{t}+\mathrm{2}\:{haven}\:{t}\:{any}\:{roots}\:\left(\:\Delta<\mathrm{0}\right)\:{so} \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{2}\right)=\mathrm{1}\:\:{and}\:\left({f}^{−\mathrm{1}} \right)^{'} \left(\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{{f}^{'} \left({f}^{−\mathrm{1}} \left(\mathrm{2}\right)\right)}=\:\frac{\mathrm{1}}{{f}^{'} \left(\mathrm{1}\right)}=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${g}^{'} \left({f}^{−\mathrm{1}} \left(\mathrm{2}\right)\right)={g}^{'} \left(\mathrm{1}\right)=\mathrm{2}\:\:{for}\:{that}\:\left({gof}^{−\mathrm{1}} \right)^{'} \left(\mathrm{2}\right)=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\left({B}\right) \\ $$