Question

Find  \[\frac{dy}{dx}\] in the following case: \[y^3 - 3x y^2 = x^3 + 3 x^2 y\] ?

 

Answer 1

\[\text{ We have }, y^3 - 3x y^2 = x^3 + 3 x^2 y\]

Differentiating with respect to x, we get,

\[\Rightarrow \frac{d}{dx}\left( y^3 \right) - \frac{d}{dx}\left( 3x y^2 \right) = \frac{d}{dx}\left( x^3 \right) + \frac{d}{dx}\left( 3 x^2 y \right)\]
\[ \Rightarrow 3 y^2 \frac{d y}{d x} - 3\left[ x\frac{d}{dx}\left( y^2 \right) + y^2 \frac{d}{dx}\left( x \right) \right] = 3 x^2 + 3\left[ x^2 \frac{d}{dx}\left( y \right) + y\frac{d}{dx}\left( x^2 \right) \right] \left[ \text{ Using product rule } \right]\]
\[ \Rightarrow 3 y^2 \frac{d y}{d x} - 3\left[ x\left( 2y \right)\frac{d y}{d x} + y^2 \right] = 3 x^2 + 3\left[ x^2 \frac{d y}{d x} + y\left( 2x \right) \right]\]
\[ \Rightarrow 3 y^2 \frac{d y}{d x} - 6xy\frac{d y}{d x} - 3 y^2 = 3 x^2 + 3 x^2 \frac{d y}{d x} + 6xy\]
\[ \Rightarrow 3 y^2 \frac{d y}{d x} - 6xy\frac{d y}{d x} - 3 x^2 \frac{d y}{d x} = 3 x^2 + 6xy + 3 y^2 \]
\[ \Rightarrow 3\frac{d y}{d x}\left( y^2 - 2xy - x^2 \right) = 3\left( x^2 + 2xy + y^2 \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{3 \left( x + y \right)^2}{3\left( y^2 - 2xy - x^2 \right)}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{\left( x + y \right)^2}{y^2 - 2xy - x^2}\]