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

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Toolbox:
  • Use chain rule for differentiating y
$\large\frac{dy}{dx}=f ' \bigg(\large \frac{2x-1}{x^2+1} \bigg) $ x $ \large\frac{d}{dx}\large\frac{2x-1}{x^2+1}$
$f '(x) = sin\:x^2$
$ \Rightarrow f ' \bigg( \large\frac{2x-1}{x^2+1} \bigg) =sin \bigg( \large\frac{2x-1}{x^2+1} \bigg)^2$
$ \large\frac{dy}{dx}=sin \bigg( \large\frac{2x-1}{x^2+1} \bigg) ^2.\large\frac{2+2x-2x^2}{(x^2+1)^2}$

 

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

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