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Find $\Large\frac{dy}{dx},$ when $y=\sin^{-1}\bigg[x\sqrt{1-x}-\sqrt x \sqrt{1-x^2}\bigg]$

1 Answer

Toolbox:
  • $ sin^{-1} ( x\sqrt{1-y^2}-y\sqrt{1-x^2})=sin^{-1}x-sin^{-1}y$
  • Take $ x=x\: and \: y=\sqrt x$ then differentiate
$ y=sin^{-1}x-sin^{-1}\sqrt x$
$ \large\frac{dy}{dx}=\large\frac{1}{\sqrt{1-x^2}}-\large\frac{1}{2\sqrt{x-x^2}}$

 

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

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