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Question Number 215809 by efronzo1 last updated on 18/Jan/25

$$\mathrm{If}\mathrm{f}\left(\mathrm{x}\right)=2+\underset{1}{\stackrel{-{\mathrm{x}}^3}{\int }}\sqrt{2+{\mathrm{u}}^2}\mathrm{du}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{\mathrm{d}}{\mathrm{dx}}{\left[{\mathrm{f}}^{-1}\left(\mathrm{x}\right)\right]}_{\mathrm{x}=2}$$

Answered by mr W last updated on 19/Jan/25

$$f\left(x\right)=2+{\int }_1^{-x^3}\sqrt{2+u^2}du$$$$\frac{df\left(x\right)}{dx}=-3x^2\sqrt{2+x^6}$$$$f(-1)=2\Rightarrow f^{-1}\left(2\right)=-1$$$$\frac{d}{dx}{\left[f^{-1}\left(x\right)\right]}_{x=2}=\frac{1}{\frac{d}{dx}{\left[f\left(x\right)\right]}_{x=-1}}$$$$=-\frac{1}{3\sqrt{3}}=-\frac{\sqrt{3}}{9}✓$$

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