$If \frac{x}{x-y}=\log \left(\frac{a}{x-y}\right), \text { then } \frac{d y}{d x}=$

$If \frac{x}{x-y}=\log \left(\frac{a}{x-y}\right), \text { then } \frac{d y}{d x}=$
  1. $2+\frac{1}{y}$
  2. $\frac{2 y-x}{y}$
  3. $\frac{2 x-y}{x}$
  4. $\frac{x-2 y}{y}$

Solution

$\frac{x}{x-y}=\log a-\log (x-y)$ $\therefore \log (x-y)+\frac{x}{x-y}=\log a$ $\therefore \frac{1}{(x-y)}\left(1-\frac{d y}{d x}\right)+\left[\frac{(x-y)(1)-(x)\left(1-\frac{d y}{d x}\right)}{x-y}\right]=0$ $\therefore\left[\frac{1}{x-y}-\frac{1}{x-y} \frac{d y}{d x}\right]+\left[\frac{x-y-x+x \frac{d y}{d x}}{x-y}\right]=0$ $\therefore \frac{1}{x-y}-\frac{1}{x-y} \frac{d y}{d x}-\frac{y}{(x-y)^{2}}+\frac{x}{(x-y)^{2}}=0$ $\therefore \quad(x-y)-(x-y) \frac{d y}{d x}-y+x \frac{d y}{d x}=0$ $\therefore \frac{d y}{d x}=\frac{x-2 y}{-y}=\frac{2 y-x}{y}$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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