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Question Number 150590 by liberty last updated on 13/Aug/21

$$\mathrm{Let}\mathrm{g}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{function}\mathrm{of}$$$$\mathrm{f}\mathrm{and}\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{10}}{1+{\mathrm{x}}^2}.\mathrm{If}\mathrm{g}\left(2\right)=a\mathrm{then}$$$$\mathrm{g}{}^{\prime }\left(2\right)=\mathrm{\_}\mathrm{\_}$$

Answered by Olaf_Thorendsen last updated on 13/Aug/21

$$f\mathrm{o}g\left(x\right)=x$$$$g^{\prime }\left(x\right)\times f^{\prime }\mathrm{o}g\left(x\right)=1$$$$g^{\prime }\left(x\right)=\frac{1}{f^{\prime }\mathrm{o}g\left(x\right)}$$$$g^{\prime }\left(2\right)=\frac{1}{f^{\prime }\mathrm{o}g\left(2\right)}=\frac{1}{\frac{g{\left(2\right)}^{10}}{1+g{\left(2\right)}^2}}=\frac{1+a^2}{a^{10}}$$

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