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Question Number 108237 by Study last updated on 15/Aug/20

$$y=\sqrt{x^2+1}-ln(\frac{1}{x}+\sqrt{1+\frac{1}{x^2}})$$$$\frac{dy}{dx}=?$$

Answered by Dwaipayan Shikari last updated on 15/Aug/20

$$y=\sqrt{x^2+1}-log(1+\sqrt{x^2+1})+logx$$$$y=\frac{x}{\sqrt{x^2+1}}-\frac{x}{\sqrt{x^2+1}+x^2+1}+\frac{1}{x}$$

Answered by john santu last updated on 15/Aug/20

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

Commented by bemath last updated on 15/Aug/20

$$thesimpleanswer$$

Answered by mathmax by abdo last updated on 15/Aug/20

$$\mathrm{y}\left(\mathrm{x}\right)=\sqrt{{\mathrm{x}}^2+1}-\ln (\frac{1}{\mathrm{x}}+\sqrt{1+\frac{1}{{\mathrm{x}}^2}})\Rightarrow$$$${\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+1}}+\frac{{(\frac{1}{\mathrm{x}}+\sqrt{1+{\mathrm{x}}^{-2}})}^{{}^{\prime }}}{\frac{1}{\mathrm{x}}+\sqrt{1+{\mathrm{x}}^{-2}}}=\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+1}}+\frac{-\frac{1}{{\mathrm{x}}^2}+\frac{-{2\mathrm{x}}^{-3}}{2\sqrt{1+{\mathrm{x}}^{-2}}}}{\frac{1}{\mathrm{x}}+\sqrt{1+{\mathrm{x}}^{-2}}}$$$$=\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}-\frac{\frac{1}{{\mathrm{x}}^2}+\frac{1}{{\mathrm{x}}^3\sqrt{1+{\mathrm{x}}^{-2}}}}{\frac{1}{\mathrm{x}}+\sqrt{1+{\mathrm{x}}^{-2}}}=\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}-\frac{1}{{\mathrm{x}}^3}.\frac{\mathrm{x}+\frac{1}{\sqrt{1+{\mathrm{x}}^{-2}}}}{(\frac{1}{\mathrm{x}}+\sqrt{1+{\mathrm{x}}^{-2}})}$$$$=\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}-\frac{\mathrm{x}\sqrt{1+{\mathrm{x}}^{-2}}+1}{\sqrt{1+{\mathrm{x}}^{-2}}({\mathrm{x}}^2+{\mathrm{x}}^3\sqrt{1+{\mathrm{x}}^2})}$$

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