Question

Find $\frac{dy}{dx}$ in each of the following cases:

${\left(x^2+y^2\right)}^2=xy$

Answer 1

$\mathrm{We}\mathrm{have},\left(x^2+y^2\right)=xy$
Differentiating with respect to x, we get,
$$\begin{gathered} \Rightarrow \frac{d}{dx}\left[{\left(x^2+y^2\right)}^2\right]=\frac{d}{dx}\left(xy\right) \\ \Rightarrow 2\left(x^2+y^2\right)\frac{d}{dx}\left(x^2+y^2\right)=x\frac{\mathit{d}y}{\mathit{d}x}+y\frac{d}{dx}\left(x\right) \\ \Rightarrow 2\left(x^2+y^2\right)\left(2x+2y\frac{\mathit{d}y}{\mathit{d}x}\right)=x\frac{\mathit{d}y}{\mathit{d}x}+y\left(1\right) \\ \Rightarrow 4x\left(x^2+y^2\right)+4y\left(x^2+y^2\right)\frac{\mathit{d}y}{\mathit{d}x}=x\frac{\mathit{d}y}{\mathit{d}x}+y \\ \Rightarrow 4y\left(x^2+y^2\right)\frac{\mathit{d}y}{\mathit{d}x}-x\frac{\mathit{d}y}{\mathit{d}x}=y-4x\left(x^2+y^2\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}\left[4y\left(x^2+y^2\right)-x\right]=y-4x\left(x^2+y^2\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{y-4x\left(x^2+y^2\right)}{4y\left(x^2+y^2\right)-x} \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{4x\left(x^2+y^2\right)-y}{x-4y\left(x^2+y^2\right)} \end{gathered}$$