Question

Find  \[\frac{dy}{dx}\] in the following case \[\left( x^2 + y^2 \right)^2 = xy\] ?

 

Answer 1

\[\text{We have, } \left( x^2 + y^2 \right) = xy\]

Differentiating with respect to x, we get,

\[\Rightarrow \frac{d}{dx}\left[ \left( x^2 + y^2 \right)^2 \right] = \frac{d}{dx}\left( xy \right)\]
\[ \Rightarrow 2\left( x^2 + y^2 \right)\frac{d}{dx}\left( x^2 + y^2 \right) = x\frac{d y}{d x} + y\frac{d}{dx}\left( x \right) \]
\[ \Rightarrow 2\left( x^2 + y^2 \right)\left( 2x + 2y\frac{d y}{d x} \right) = x\frac{d y}{d x} + y\left( 1 \right)\]
\[ \Rightarrow 4x\left( x^2 + y^2 \right) + 4y\left( x^2 + y^2 \right)\frac{d y}{d x} = x\frac{d y}{d x} + y\]
\[ \Rightarrow 4y\left( x^2 + y^2 \right)\frac{d y}{d x} - x\frac{d y}{d x} = y - 4x\left( x^2 + y^2 \right)\]
\[ \Rightarrow \frac{d y}{d x}\left[ 4y\left( x^2 + y^2 \right) - x \right] = y - 4x\left( x^2 + y^2 \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{y - 4x\left( x^2 + y^2 \right)}{4y\left( x^2 + y^2 \right) - x}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{4x\left( x^2 + y^2 \right) - y}{x - 4y\left( x^2 + y^2 \right)}\]