Question

If $x^{16}{y}^9={\left(x^2+y\right)}^{17}$, prove that $x\frac{dy}{dx}=2y$

Answer 1

$\mathrm{We}\mathrm{have},x^{16}{y}^9={\left(x^2+y\right)}^{17}$
Taking log on both sides,
$$\begin{gathered} \log \left(x^{16}{y}^9\right)=\log {\left(x^2+y\right)}^{17} \\ \Rightarrow 16\log x+9\log y=17\log \left(x^2+y\right) \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} 16\frac{d}{dx}\left(\log x\right)+9\frac{d}{dx}\left(\log y\right)=17\frac{d}{dx}\log \left(x^2+y\right) \\ \Rightarrow \frac{16}{x}+\frac{9}{y}\frac{dy}{dx}=\frac{17}{x^2+y}\frac{d}{dx}\left(x^2+y\right) \\ \Rightarrow \frac{16}{x}+\frac{9}{y}\frac{dy}{dx}=\frac{17}{x^2+y}\left[2x+\frac{dy}{dx}\right] \\ \Rightarrow \frac{9}{y}\frac{dy}{dx}-\frac{17}{x^2+y}\frac{dy}{dx}=\frac{34x}{x^2+y}-\frac{16}{x} \\ \Rightarrow \frac{dy}{dx}\left[\frac{9}{y}-\frac{17}{x^2+y}\right]=\frac{34x}{x^2+y}-\frac{16}{x} \\ \Rightarrow \frac{dy}{dx}\left[\frac{9\left(x^2+y\right)-17y}{y\left(x^2+y\right)}\right]=\left[\frac{34x^2-16\left(x^2+y\right)}{x\left(x^2+y\right)}\right] \\ \Rightarrow \frac{dy}{dx}\left[\frac{9x^2+9y-17y}{y\left(x^2+y\right)}\right]=\left[\frac{34x^2-16x^2-16y}{x\left(x^2+y\right)}\right] \\ \Rightarrow \frac{dy}{dx}\left[\frac{9x^2-8y}{y\left(x^2+y\right)}\right]=\left[\frac{18x^2-16y}{x\left(x^2+y\right)}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}\left[\frac{2\left(9x^2-8y\right)}{9x^2-8y}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{2y}{x} \\ \Rightarrow x\frac{dy}{dx}=2y \end{gathered}$$