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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

$$wehavef\left(x\right)=x^3+xandg\left(x\right)=x^3-x\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=3x^2+1and{g}^{{}^{\prime }}\left(x\right)=3x^2-1and$$$${\left({gof}^{-1}\right)}^{{}^{\prime }}\left(2\right)=g^{{}^{\prime }}\left(f^{-1}\left(2\right)\right).f^{-1^{{}^{\prime }}}\left(2\right)but{f}^{-1}\left(2\right)=t\iff f\left(t\right)=2$$$$\iff t^3+t-2=0\iff (t-1)(t^2+t+2)=0\iff t=1because$$$$thepolynomial{t}^2+t+2haventanyroots(\mathrm{\Delta }<0)so$$$$f^{-1}\left(2\right)=1and{\left(f^{-1}\right)}^{{}^{\prime }}\left(2\right)=\frac{1}{f^{{}^{\prime }}\left(f^{-1}\left(2\right)\right)}=\frac{1}{f^{{}^{\prime }}\left(1\right)}=\frac{1}{4}$$$$g^{{}^{\prime }}\left(f^{-1}\left(2\right)\right)=g^{{}^{\prime }}\left(1\right)=2forthat{\left({gof}^{-1}\right)}^{{}^{\prime }}\left(2\right)=2\times \frac{1}{4}=\frac{1}{2}\left(B\right)$$

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