Question

If u=log(x/y) then the value of a
x du/dx + y du/dy is equal to​

Answer 1

Answer:

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

Explanation:

$u = \:log ( \frac{x}{y} ) \\ u= logx - logy \\ \frac{du}{dx} = \frac{d}{dx} (logx - logy) \\ = \frac{d}{dx} (logx) - \frac{d}{dx} (logy) \\ = \frac{1}{x} + \frac{1}{y} \frac{dy}{dx}$

$\frac{du}{dx} = \frac{1}{x} + \frac{1}{y} \frac{dy}{dx}$

$\frac{du}{dy} = \frac{d}{dy} (logx - logy) \\ = \frac{d}{dy}(logx) - \frac{d}{dy} (logy) \\ = \frac{1}{x} \frac{dx}{dy} - \frac{1}{y}$

$\frac{du}{dy} = \frac{1}{x} \frac{dx}{dy} - \frac{1}{y}$

$x \frac{du}{dx} + y \frac{du}{dy} = x(\frac{1}{x} + \frac{1}{y} \frac{dy}{dx} ) + y(\frac{1}{x} \frac{dx}{dy} + \frac{1}{y}) \\ = 1 + \frac{x}{y} \frac{dy}{dx} + \frac{y}{x} \frac{dx}{dy} + 1 \\ = 2 + \frac{x}{y} \frac{dy}{dx} + \frac{y}{x} \frac{dx}{dy}$

Answer 2

Explanation:

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