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Question Number 106039 by 175mohamed last updated on 02/Aug/20

$$if\left(fog\right)\left(x\right)=xand{f}^{\prime }\left(x\right)=1+{\left(f\left(x\right)\right)}^2$$$$then{g}^{\prime }\left(2\right)=....$$$$$$

Answered by bemath last updated on 02/Aug/20

$$\frac{\mathrm{d}(\mathrm{f}\circ \mathrm{g})\left(\mathrm{x}\right)}{\mathrm{dx}}={\mathrm{g}}^{\prime }\left(\mathrm{x}\right).\mathrm{f}{}^{\prime }\left(\mathrm{g}\left(\mathrm{x}\right)\right)=1$$$${\mathrm{g}}^{\prime }\left(2\right).\mathrm{f}{}^{\prime }\left(\mathrm{g}\left(2\right)\right)=1$$$$\mathrm{g}{}^{\prime }\left(2\right)=\frac{1}{1+(\mathrm{f}{\left(\mathrm{g}\left(2\right)\right)}^2}$$$$\iff \mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=\mathrm{x}\Rightarrow \mathrm{g}\left(\mathrm{x}\right)={\mathrm{f}}^{-1}\left(\mathrm{x}\right)$$$$\mathrm{so}\mathrm{g}\left(2\right)={\mathrm{f}}^{-1}\left(2\right)$$$$\therefore \mathrm{g}{}^{\prime }\left(2\right)=\frac{1}{1+(\mathrm{f}{\left({\mathrm{f}}^{-1}\left(2\right)\right)}^2}=\frac{1}{1+4}=\frac{1}{5}$$

Answered by maouame last updated on 02/Aug/20

$$if\left(fog\right)\left(x\right)=xand{f}^{\prime }\left(x\right)=1+{\left(f\left(x\right)\right)}^2$$$$then{g}^{\prime }\left(2\right)=....$$$$Ona{\left(fog\right)}^{\prime }\left(x\right)=g^{\prime }\left(x\right)\times f^{\prime }\left(g\left(x\right)\right)$$$$Alors{g}^{\prime }\left(x\right)\times f^{\prime }\left(g\left(x\right)\right)=1$$$$\Rightarrow g^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(g\left(x\right)\right)}(car{f}^{\prime }\left(x\right)>0)$$$$\Rightarrow g^{\prime }\left(x\right)=\frac{1}{1+{\left(f\left(g\left(x\right)\right)\right)}^2}$$$$\Rightarrow g^{\prime }\left(x\right)=\frac{1}{1+x^2}$$$$\Rightarrow g^{\prime }\left(2\right)=\frac{1}{5}$$

Answered by mathmax by abdo last updated on 02/Aug/20

$$\mathrm{fog}\left(\mathrm{x}\right)=\mathrm{x}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{g}\left(\mathrm{x}\right)\right).{\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)=1\mathrm{but}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=1+{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=1+{\left(\mathrm{fog}\left(\mathrm{x}\right)\right)}^2=1+{\mathrm{x}}^2\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{g}\left(2\right)\right)=1+2^2=5\mathrm{and}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{g}\left(2\right)\right).{\mathrm{g}}^{{}^{\prime }}\left(2\right)=1\Rightarrow {\mathrm{g}}^{{}^{\prime }}\left(2\right)=\frac{1}{5}$$

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