Question

If \[xy = 1\] prove that \[\frac{dy}{dx} + y^2 = 0\] ?

Answer 1

We have,

`xy = 1`

Differentiating with respect to x, we get, 

\[\frac{d}{dx}\left( xy \right) = \frac{d}{dx}\left( 1 \right)\]

\[ \Rightarrow x\frac{d y}{d x} + y\frac{d}{dx}\left( x \right) = 0 ..........\left[ \text{ Using product rule} \right]\]

\[ \Rightarrow x\frac{d y}{d x} + y\left( 1 \right) = 0\]

\[ \Rightarrow \frac{d y}{d x} = - \frac{y}{x} \]

\[ \Rightarrow \frac{d y}{d x} = - \frac{y}{\frac{1}{y}} ..........\left[ \because x = \frac{1}{y} \right]\]

\[ \Rightarrow \frac{d y}{d x} = - y^2 \]

\[ \Rightarrow \frac{d y}{d x} + y^2 = 0\]