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If $ y=\sqrt{x^2-1} - log \bigg(\Large\frac{1}{x}+\sqrt{1+\frac{1}{x^2}}\bigg),\;\normalsize find\;\Large\frac{dy}{dx}$

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Toolbox:
  • Use chain rule and differentiate
$\large \frac{dy}{dx}=\large\frac{2x}{2\sqrt{x^2-1}}- \bigg[\large\frac{1}{\large\frac{1}{x}+\sqrt{1+\large\frac{1}{x^2}}} \bigg]$ x $ \bigg( \large\frac{-1}{x^2}+\large\frac{1}{2\sqrt{1+\large\frac{1}{x^2}}} $ x $ \bigg(\large \frac{-2}{x^3} \bigg) \bigg )$
 
Simplify and get the result
$ \large\frac{dy}{dx}=\large\frac{x^2\sqrt{x^2+1}+\sqrt{x^2-1}}{x\sqrt{x^4-1}} $

 

answered Mar 9, 2013 by thanvigandhi_1
edited Mar 25, 2013 by thanvigandhi_1
 

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