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Question Number 135313 by bobhans last updated on 12/Mar/21

$$Givenf\left(x\right)=5x+\cos \left(3x\right)$$$$Findthevalueof\frac{d}{dx}\left[f^{-1}\left(1\right)\right]$$

Answered by liberty last updated on 12/Mar/21

$$\left(1\right)f\left(0\right)=1\iff f^{-1}\left(1\right)=0$$$$\left(2\right)\frac{d}{dx}\left[f^{-1}\left(1\right)\right]=\frac{1}{5-3\sin 3x}{\mid }_{x=f^{-1}\left(1\right)}$$$$=\frac{1}{5-3\sin 3x}{\mid }_{x=0}=\frac{1}{5}$$

Answered by mathmax by abdo last updated on 12/Mar/21

$${\mathrm{we}\mathrm{know}{\left({\mathrm{f}}^{-1}\left({\mathrm{x}}_0\right)\right)}^{{}^{\prime }}=\frac{1}{{\mathrm{f}}^{{}^{\prime }}\left({\mathrm{f}}^{-1}\left({\mathrm{x}}_0\right)\right)}\Rightarrow {\mathrm{f}}^{-1}\left(1\right))}^{{}^{\prime }}=\frac{1}{{\mathrm{f}}^{{}^{\prime }}\left({\mathrm{f}}^{-1}\left(1\right)\right)}$$$$\mathrm{f}\left(0\right)=1\Rightarrow {\mathrm{f}}^{-1}\left(1\right)=0\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{f}}^{-1}\left(1\right)\right)=\frac{1}{{\mathrm{f}}^{{}^{\prime }}\left(0\right)}=\frac{1}{5}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=5-3\mathrm{sinx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(0\right)=5$$

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