Question

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

$y^3-3x{y}^2=x^3+3x^{2}y$

Answer 1

$\mathrm{We}\mathrm{have},y^3-3x{y}^2=x^3+3x^{2}y$
Differentiating with respect to x, we get,
$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(y^3\right)-\frac{d}{dx}\left(3x{y}^2\right)=\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(3x^{2}y\right) \\ \Rightarrow 3y^2\frac{\mathit{d}y}{\mathit{d}x}-3\left[x\frac{d}{dx}\left(y^2\right)+y^2\frac{d}{dx}\left(x\right)\right]=3x^2+3\left[x^2\frac{d}{dx}\left(y\right)+y\frac{d}{dx}\left(x^2\right)\right]\left[\mathrm{Using}\mathrm{product}\mathrm{rule}\right] \\ \Rightarrow 3y^2\frac{\mathit{d}y}{\mathit{d}x}-3\left[x\left(2y\right)\frac{\mathit{d}y}{\mathit{d}x}+y^2\right]=3x^2+3\left[x^2\frac{\mathit{d}y}{\mathit{d}x}+y\left(2x\right)\right] \\ \Rightarrow 3y^2\frac{\mathit{d}y}{\mathit{d}x}-6xy\frac{\mathit{d}y}{\mathit{d}x}-3y^2=3x^2+3x^2\frac{\mathit{d}y}{\mathit{d}x}+6xy \\ \Rightarrow 3y^2\frac{\mathit{d}y}{\mathit{d}x}-6xy\frac{\mathit{d}y}{\mathit{d}x}-3x^2\frac{\mathit{d}y}{\mathit{d}x}=3x^2+6xy+3y^2 \\ \Rightarrow 3\frac{\mathit{d}y}{\mathit{d}x}\left(y^2-2xy-x^2\right)=3\left(x^2+2xy+y^2\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{3{\left(x+y\right)}^2}{3\left(y^2-2xy-x^2\right)} \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{{\left(x+y\right)}^2}{y^2-2xy-x^2} \end{gathered}$$