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Question:

If $x^{m} y^{n}=1$, prove that $\frac{d y}{d x}=-\frac{m y}{n x}$

Solution:

Here,

$x^{m} y^{n}=1$

Taking log on both sides,

$\log \left(x^{m} y^{n}\right)=\log 1$

$m \log x+n \log y=\log 1\left[\right.$ Since, $\left.\log (A B)=\log A+\log B ; \log a^{b}=b \log a\right]$

Differentiating with respect to $x$

$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{m} \log \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{nlogy})=\frac{\mathrm{d}}{\mathrm{dx}}(\log (1))$

$\frac{\mathrm{m}}{\mathrm{x}}+\frac{\mathrm{n}}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=0$

$\frac{\mathrm{n}}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{m}}{\mathrm{x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{m}}{\mathrm{x}} \times \frac{\mathrm{y}}{\mathrm{n}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{my}}{\mathrm{nx}}$

Hence Proved.

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