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Find$ \Large\frac{dy}{dx}\;\normalsize when\;y=\sin ^{-1} \bigg[\Large\frac{5x+12 \sqrt{1-x^2}}{13}\bigg]$

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  • \( sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2} )=sin^{-1}x+sin^{-1}y\)
Take x as x and y as $ \large\frac{12}{13}$ in the formula
$ y=sin^{-1} \bigg[ x \sqrt{1-\large\frac{144}{169}}+\large\frac{12}{13}\sqrt{1-x^2} \bigg]$
$ = sin^{-1}x+sin^{-1}\large\frac{12}{13}$
 
$ \large\frac{dy}{dx}=\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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