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If $x^{2} y^{b}=(x-y)^{a+b}$, then the value of $\frac{d y}{d x}-\frac{y}{x}$ is equal to
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$x^{a} y^{b}=(x-y)^{a+b}$
taking log both the sides.
$\log \left(x^{a} y^{b}\right)=\log (x-y)^{(a+b)}$
$a \log x+b \log y=(a+b) \log (x-y)$
differentiating both sides w.r.t 'x'.
$\frac{a}{x}+\frac{b}{y} \frac{d y}{d x}=\frac{(a+b)}{(x-y)}\left[1-\frac{d y}{d x}\right]$
$\frac{d y}{d x}\left[\frac{b}{y}+\frac{a+b}{x-y}\right]=\frac{a+b}{x-y}-\frac{a}{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{\mathrm{bx}-\mathrm{by}+\mathrm{ay}+\mathrm{by}}{\mathrm{y}(\mathrm{x}-\mathrm{y})}\right]=\frac{\mathrm{ax}+\mathrm{bx}-\mathrm{ax}+\mathrm{ay}}{\mathrm{x}(\mathrm{x}-\mathrm{y})}$
$\frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{\mathrm{bx}+\mathrm{ay}}{\mathrm{y}}\right]=\frac{\mathrm{bx}+\mathrm{ay}}{\mathrm{x}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=0$
taking log both the sides.
$\log \left(x^{a} y^{b}\right)=\log (x-y)^{(a+b)}$
$a \log x+b \log y=(a+b) \log (x-y)$
differentiating both sides w.r.t 'x'.
$\frac{a}{x}+\frac{b}{y} \frac{d y}{d x}=\frac{(a+b)}{(x-y)}\left[1-\frac{d y}{d x}\right]$
$\frac{d y}{d x}\left[\frac{b}{y}+\frac{a+b}{x-y}\right]=\frac{a+b}{x-y}-\frac{a}{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{\mathrm{bx}-\mathrm{by}+\mathrm{ay}+\mathrm{by}}{\mathrm{y}(\mathrm{x}-\mathrm{y})}\right]=\frac{\mathrm{ax}+\mathrm{bx}-\mathrm{ax}+\mathrm{ay}}{\mathrm{x}(\mathrm{x}-\mathrm{y})}$
$\frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{\mathrm{bx}+\mathrm{ay}}{\mathrm{y}}\right]=\frac{\mathrm{bx}+\mathrm{ay}}{\mathrm{x}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=0$
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