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

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.
$y=sin^{-1}x^2+sin^{-1}x$

$\large\frac{dy}{dx}=\large\frac{2x}{\sqrt{1-x^4}}+\large\frac{1}{\sqrt{1-x^2}}$

edited Mar 26, 2013