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Question Number 153963 by ZiYangLee last updated on 12/Sep/21

$$\mathrm{Given}\mathrm{that}f\mathrm{and}g\mathrm{are}\mathrm{differentiable}$$$$\mathrm{functions}\mathrm{such}\mathrm{that}f{}^{\prime }\left(x\right)=\frac{1}{\sqrt{1+{\left(f\left(x\right)\right)}^2}},$$$$\mathrm{and}g=f^{-1},\mathrm{find}{g}^{\prime }\left(x\right).$$

Answered by mr W last updated on 12/Sep/21

$$y=f\left(x\right)$$$$y^{\prime }=\frac{1}{\sqrt{1+y^2}}$$$$\sqrt{1+y^2}dy=dx$$$$\frac{1}{2}[\ln (y+\sqrt{1+y^2})+y\sqrt{1+y^2}]+C=x$$$$\Rightarrow g\left(x\right)=f^{-1}\left(x\right)=\frac{1}{2}[\ln (x+\sqrt{1+x^2})+x\sqrt{1+x^2}]+C$$$$g^{\prime }\left(x\right)=\frac{1}{2}[\frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}+\sqrt{1+x^2}+\frac{x^2}{\sqrt{1+x^2}}]$$$$=\frac{1}{2}[\frac{x+\sqrt{1+x^2}}{x+\sqrt{1+x^2}}+1+2x^2]\frac{1}{\sqrt{1+x^2}}$$$$=\sqrt{1+x^2}$$

Commented by ZiYangLee last updated on 12/Sep/21

$$\mathrm{i}\mathrm{found}\mathrm{a}\mathrm{better}\mathrm{way}..$$

Commented by mr W last updated on 12/Sep/21

$$yes,yourwayisbest!$$

Answered by ZiYangLee last updated on 12/Sep/21

$$letg\left(x\right)=f^{-1}\left(x\right)=y$$$$f\left(y\right)=x$$$$f{}^{\prime }\left(x\right)\frac{dy}{dx}=1$$$$\frac{1}{\sqrt{1+{\left[f\left(y\right)\right]}^2}}\frac{dy}{dx}=1$$$$\frac{dy}{dx}=\sqrt{1+{\left[f\left(y\right)\right]}^2}$$$$g{}^{\prime }\left(x\right)=\sqrt{1+{\left[f\left(y\right)\right]}^2}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

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