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If x3+y3=3axy, then find dy/dx.

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Given- $x^3 + y^3 = 3a x y$

To find- $\dfrac{dy}{dx}$

Solution- $x^3 + y^3 = 3a x y$
Differentiate both sides with respect to x:
$\dfrac{d}{dx}(x^3) + \dfrac{d}{dx}(y^3) = \dfrac{d}{dx}(3a x y)$
Apply the rules:
$3x^2 + 3y^2 \dfrac{dy}{dx} = 3a\left( x \dfrac{dy}{dx} + y \right)$
Divide both sides by 3:
$x^2 + y^2 \dfrac{dy}{dx} = a\left( x \dfrac{dy}{dx} + y \right)$

Now, bring terms involving $\dfrac{dy}{dx}$ to one side:

$y^2 \dfrac{dy}{dx} - a x \dfrac{dy}{dx} = a y - x^2$
$\left( y^2 - a x \right) \dfrac{dy}{dx} = a y - x^2$
$\dfrac{dy}{dx} = \dfrac{a y - x^2}{y^2 - a x}$

Posted by

Saniya Khatri

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