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If $y=\sin ^{-1}\big [x^2\sqrt{1-x^2}+x\sqrt {1-x^4}\big],$ prove that $ \Large\frac{dy}{dx}=\frac{2x}{\sqrt{1-x^4}}+\frac{1}{\sqrt{1-x^2}}$

1 Answer

Toolbox:
  • $ sin^{-1} [ x\sqrt{1-y^2}+y\sqrt{1-x^2} ] = sin^{-1}x+sin^{-1}y$
  • Consider $ x \: as \: x^2\: and \: y\: as \: x$ in the formula.
$ y=sin^{-1}x^2+sin^{-1}x$
 
$ \large\frac{dy}{dx}=\large\frac{2x}{\sqrt{1-x^4}}+\large\frac{1}{\sqrt{1-x^2}}$

 

answered Mar 9, 2013 by thanvigandhi_1
edited Mar 26, 2013 by thanvigandhi_1
 

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