# #x*y = a(y+√(y²-x²))# then show that #x³*dy/dx=y²(y+√(y²-x²))# ?

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To show that ( x^3 \frac{dy}{dx} = y^2(y + \sqrt{y^2 - x^2}) ), differentiate both sides of the given equation ( xy = a(y + \sqrt{y^2 - x^2}) ) with respect to ( x ) implicitly, then manipulate the resulting expression to obtain the desired form.

First, differentiate both sides of the equation ( xy = a(y + \sqrt{y^2 - x^2}) ) with respect to ( x ) using the product rule and the chain rule:

[ \frac{d}{dx}(xy) = \frac{d}{dx}(a(y + \sqrt{y^2 - x^2})) ]

[ y + x\frac{dy}{dx} = a\left(\frac{dy}{dx} + \frac{x}{\sqrt{y^2 - x^2}}\right) ]

[ y + x\frac{dy}{dx} = a\frac{dy}{dx} + \frac{ax}{\sqrt{y^2 - x^2}} ]

Now, solve for ( \frac{dy}{dx} ):

[ x\frac{dy}{dx} - a\frac{dy}{dx} = a\frac{ax}{\sqrt{y^2 - x^2}} - y ]

[ (x - a)\frac{dy}{dx} = \frac{ax}{\sqrt{y^2 - x^2}} - y ]

[ \frac{dy}{dx} = \frac{ax}{(x - a)\sqrt{y^2 - x^2}} - \frac{y}{x - a} ]

Now, multiply both sides by ( x^3 ):

[ x^3 \frac{dy}{dx} = x^3\left(\frac{ax}{(x - a)\sqrt{y^2 - x^2}} - \frac{y}{x - a}\right) ]

[ x^3 \frac{dy}{dx} = \frac{ax^4}{(x - a)\sqrt{y^2 - x^2}} - \frac{x^3y}{x - a} ]

[ x^3 \frac{dy}{dx} = \frac{ax^4}{(x - a)\sqrt{y^2 - x^2}} - \frac{x^3y - ax^3}{x - a} ]

[ x^3 \frac{dy}{dx} = \frac{ax^4}{(x - a)\sqrt{y^2 - x^2}} - \frac{ax^3}{x - a} ]

[ x^3 \frac{dy}{dx} = \frac{ax^4 - ax^3\sqrt{y^2 - x^2}}{(x - a)\sqrt{y^2 - x^2}} ]

[ x^3 \frac{dy}{dx} = \frac{ax^3(x - \sqrt{y^2 - x^2})}{(x - a)\sqrt{y^2 - x^2}} ]

[ x^3 \frac{dy}{dx} = \frac{ax^3(y + \sqrt{y^2 - x^2})}{(x - a)\sqrt{y^2 - x^2}} ]

[ x^3 \frac{dy}{dx} = y^2(y + \sqrt{y^2 - x^2}) ]

Thus, we have shown that ( x^3 \frac{dy}{dx} = y^2(y + \sqrt{y^2 - x^2}) ).

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