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If $x^{m}+y^{m}=1$ such that $\frac{d y}{d x}=-\frac{x}{y}$, then what should be the value of $\mathrm{m} ?$
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2
Let $x^{\mathrm{m}}+\mathrm{y}^{\mathrm{m}}=1$
Differentiate both the sides w.r.t $^{\prime} x^{\prime}$
$\mathrm{m} \cdot \mathrm{x}^{\mathrm{m}-1}+\mathrm{m} \cdot \mathrm{y}^{\mathrm{m}-1} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=0$
$\Rightarrow \quad \mathrm{m} \mathrm{x}^{\mathrm{m}-1}=-\mathrm{m} \mathrm{y}^{\mathrm{m}-1} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
$\Rightarrow \frac{\mathrm{x}^{\mathrm{m}-1}}{\mathrm{y}^{\mathrm{m}-1}}=-\frac{\mathrm{dy}}{\mathrm{dx}}$
$\Rightarrow \left(\frac{x}{y}\right)^{m-1}=\frac{x}{y} \quad\left(\because \frac{d y}{d x}=-\frac{x}{y}\right)$
$\Rightarrow \quad m-1=1 \Rightarrow m=2$
Differentiate both the sides w.r.t $^{\prime} x^{\prime}$
$\mathrm{m} \cdot \mathrm{x}^{\mathrm{m}-1}+\mathrm{m} \cdot \mathrm{y}^{\mathrm{m}-1} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=0$
$\Rightarrow \quad \mathrm{m} \mathrm{x}^{\mathrm{m}-1}=-\mathrm{m} \mathrm{y}^{\mathrm{m}-1} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
$\Rightarrow \frac{\mathrm{x}^{\mathrm{m}-1}}{\mathrm{y}^{\mathrm{m}-1}}=-\frac{\mathrm{dy}}{\mathrm{dx}}$
$\Rightarrow \left(\frac{x}{y}\right)^{m-1}=\frac{x}{y} \quad\left(\because \frac{d y}{d x}=-\frac{x}{y}\right)$
$\Rightarrow \quad m-1=1 \Rightarrow m=2$
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