Question

If $x^{13}{y}^7={\left(x+y\right)}^{20}$, prove that $\frac{dy}{dx}=\frac{y}{x}$

Answer 1

$\mathrm{We}\mathrm{have},x^{13}{y}^7={\left(x+y\right)}^{20}$
Taking log on both sides,
$$\begin{gathered} \log \left(x^{13}{y}^7\right)=\log {\left(x+y\right)}^{20} \\ \Rightarrow 13\log x+7\log y=20\log \left(x+y\right) \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} 13\frac{d}{dx}\left(\log x\right)+7\frac{d}{dx}\left(\log y\right)=20\frac{d}{dx}\log \left(x+y\right) \\ \Rightarrow \frac{13}{x}+\frac{7}{y}\frac{dy}{dx}=\frac{20}{x+y}\frac{d}{dx}\left(x+y\right) \\ \Rightarrow \frac{13}{x}+\frac{7}{y}\frac{dy}{dx}=\frac{20}{x+y}\left[1+\frac{dy}{dx}\right] \\ \Rightarrow \frac{7}{y}\frac{dy}{dx}-\frac{20}{x+y}\frac{dy}{dx}=\frac{20}{x+y}-\frac{13}{x} \\ \Rightarrow \frac{dy}{dx}\left[\frac{7}{y}-\frac{20}{x+y}\right]=\frac{20}{x+y}-\frac{13}{x} \\ \Rightarrow \frac{dy}{dx}\left[\frac{7\left(x+y\right)-20y}{y\left(x+y\right)}\right]=\left[\frac{20x-13\left(x+y\right)}{x\left(x+y\right)}\right] \\ \Rightarrow \frac{dy}{dx}=\left[\frac{20x-13x-13y}{x\left(x+y\right)}\right]\left[\frac{y\left(x+y\right)}{7x+7y-20y}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}\left(\frac{7x-13y}{7x-13y}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x} \end{gathered}$$