Question

Find $\frac{dy}{dx}$ when 𝑥 and 𝑦 are connected by the relation $(x^2+y^2{)}^2=xy$

Answer 1

Given:$(x^2+y^2{)}^2=xy$
$\therefore \frac{d}{dx}{(x^2+y^2)}^2=\frac{d}{dx}\left(xy\right)$
$\Rightarrow 2(x^2+y^2)\frac{d}{dx}(x^2+y^2)=y+x\frac{dy}{dx}$
$[\because \frac{d\left(uv\right)}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}]$
$\Rightarrow 2.(x^2+y^2).(2x+2y\frac{dy}{dx})=y+x\frac{dy}{dx}$
$\Rightarrow 4x.(x^2+y^2)-y=[x-4y.(x^2+y^2)]\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{4x_{.}(x^2+y^2)-y}{x-4y.(x^2+y^2)}$
$\Rightarrow \frac{dy}{dx}=\frac{y-4x^3-4x{y}^2}{4x^{2}y+4y^3-x}$
Hence, the value of
$\frac{dy}{dx}\text{is}\frac{y-4x^3-4x{y}^2}{4x^{2}y+4y^3-x}$