Question

If $xy=1$, prove that $\frac{dy}{dx}+y^2=0$

Answer 1

$\mathrm{We}\mathrm{have},xy=1$
Differentiating with respect to x, we get,
$$\begin{gathered} \frac{d}{dx}\left(xy\right)=\frac{d}{dx}\left(1\right) \\ \Rightarrow x\frac{\mathit{d}y}{\mathit{d}x}+y\frac{d}{dx}\left(x\right)=0\left[\mathrm{Using}\mathrm{product}\mathrm{rule}\right] \\ \Rightarrow x\frac{\mathit{d}y}{\mathit{d}x}+y\left(1\right)=0 \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=-\frac{y}{x} \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=-\frac{y}{{\displaystyle \frac{1}{y}}}\left[\because x=\frac{1}{y}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=-y^2 \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}+y^2=0 \end{gathered}$$