Question

If $x^m{y}^n=1$, prove that $\frac{dy}{dx}=-\frac{my}{nx}$

Answer 1

$\mathrm{We}\mathrm{have},x^m{y}^n=1$
Taking log on both side,
$$\begin{gathered} \log \left(x^m{y}^n\right)=\log \left(1\right) \\ \Rightarrow m\log x+n\log y=\log \left(1\right) \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \frac{dy}{dx}\left(m\log x\right)+\frac{d}{dx}\left(n\log y\right)=\frac{d}{dx}\left\{\log \left(1\right)\right\} \\ \Rightarrow \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=-\frac{m}{x}\times \frac{y}{n} \\ \Rightarrow \frac{dy}{dx}=-\frac{my}{nx} \end{gathered}$$