Question

If $xy=4$, prove that $x\left(\frac{dy}{dx}+y^2\right)=3y$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},xy=4 \\ \Rightarrow y=\frac{4}{x} \end{gathered}$$
Differentiate it with respect to x,
$$\begin{gathered} \frac{\mathit{d}y}{\mathit{d}x}=\frac{d}{dx}\left(\frac{4}{x}\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=4\frac{d}{dx}\left(x^{-1}\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=4\left(-1\times x^{-1-1}\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=4\left(-\frac{1}{x^2}\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{-4}{x^2} \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=-\frac{4}{{\left({\displaystyle \frac{4}{y}}\right)}^2}\left[\because x=\frac{4}{y}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=-\frac{4y^2}{16} \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=-\frac{y^2}{4} \\ \Rightarrow 4\frac{\mathit{d}y}{\mathit{d}x}=-y^2 \\ \Rightarrow 4\frac{\mathit{d}y}{\mathit{d}x}=3y^2-4y^2 \\ \Rightarrow 4\frac{\mathit{d}y}{\mathit{d}x}+4y^2=3y^2 \\ \Rightarrow 4\left(\frac{\mathit{d}y}{\mathit{d}x}+y^2\right)=3y^2 \end{gathered}$$
Dividing both side by x,
$$\begin{gathered} \Rightarrow \frac{4}{x}\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+y^2\right)=\frac{3y^2}{x} \\ \Rightarrow y\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+y^2\right)=\frac{3y^2}{x} \\ \Rightarrow x\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+y^2\right)=\frac{3y^2}{y} \\ \Rightarrow x\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+y^2\right)=3y \end{gathered}$$