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If $ y=\Large\frac{1}{\sqrt{b^2-a^2}} \normalsize log \bigg[\Large\frac{\sqrt{b+a}+\sqrt{b-a}\tan \frac{x}{2}} {\sqrt{b+a}-\sqrt{b-a}\tan \frac{x}{2}}\bigg]$ prove that $ \Large\frac {dy}{dx}=\frac{\sec^2\frac{x}{2}}{(b+a)-(b-a) \tan^2\frac{x}{2}}$

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Toolbox:
  • Differentiate using chain rule and quotient rule.
$ \large\frac{dy}{dx}=\bigg( \large\frac{1}{\sqrt{b^2-a^2}} \bigg)$x $ \bigg(\large \frac{\sqrt{b+a}+\sqrt{b-a}tan\large\frac{x}{2}}{\sqrt{b+a}+\sqrt{b-a}tan\large\frac{x}{2}} \bigg)$ x$ \large\frac{d}{dx} \bigg( \large\frac{(\sqrt{b+a}+\sqrt{b-a}tan\large\frac{x}{2})}{(\sqrt{b+a}-\sqrt{b-a}tan\large\frac{x}{2})} \bigg)$
 
Simplify and get
$ \large\frac{dy}{dx}=\large\frac{sec^2\large\frac{x}{2}}{(b+a)-(b-a)tan^2\large\frac{x}{2}}$

 

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

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