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If $x^2+y^2=t-\large\frac{1}{t} ,$$x^4+y^4 =t^2+\large\frac{1}{t^2}$ then $\large\frac{dy}{dx}$

$(a)\;\frac{1}{xy^3} \\ (b)\;\frac{1}{x^3y} \\ (c)\;\frac{-1}{xy^3} \\ (d)\;\frac{-1}{x^3y} $
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$x^2+y^2=t-\large\frac{1}{t} $
=> $ x^4 +y^4 +2x^2y^2=t^2 +\large\frac{1}{t^2}$$-2$
=> $x^4+y^4 +2x^2y^2 =x^4+y^4-2$
$x^2y^2= -1$
$y^2=\large\frac{-1}{x^2} $
$2y \large\frac{dy}{dx}$$ =-(-2) \large\frac{1}{x^3}$
$\large\frac{dy}{dx} =\frac{1}{x^3y}$
Hence b is the correct answer
answered Jan 31, 2014 by meena.p
 

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