Question

if xy =1 then y^2 + dy/dx =​

Answer 1

Answer:

$answer \: is \: y(y - \frac{1}{x}) = 0$

$\frac{1}{x} ( \frac{1}{x} - \frac{1}{x}) = 0$

Step-by-step explanation:

$differentiate wrt \: x$

$xy = 1 \\ y + x \times \frac{dy}{dx } = 0 \\ \frac{dy}{dx } = - \frac{y}{x}$

${y}^{2} + \frac{dy}{dx} = {y}^{2} - \frac{y}{x}$

$= y(y - \frac{1}{x} )$