Question

If y = x log y show that x dy/dx = y^2/(y - x)

Answer 1

Step-by-step explanation:

We have,

$y = x log(y)$

$\frac{d}{dx} (y)= log(y) \frac{d}{dx}(x) + x \frac{d}{dx}( log(y) )$

$\implies \frac{dy}{dx}= log(y) + x \frac{1}{y} . \frac{dy}{dx}$

$\implies \frac{dy}{dx}= \frac{y}{x} + \frac{x}{y} . \frac{dy}{dx}$

$\implies \frac{dy}{dx } - \frac{x}{y} . \frac{dy}{dx} = \frac{y}{x}$

$\implies (1 - \frac{x}{y}) . \frac{dy}{dx} = \frac{y}{x}$

$\implies ( \frac{y - x}{y}) . \frac{dy}{dx} = \frac{y}{x}$

$\implies \frac{dy}{dx} = \frac{ {y}^{2} }{x(y - x)}$

$\implies x\frac{dy}{dx} = \frac{ {y}^{2} }{(y - x)}$