# Solve this

Question:

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

Solution:

Here,

$x^{16} y^{9}=\left(x^{2}+y\right)^{17}$

Taking log on both sides,

$\log \left(x^{16} y^{9}\right)=\log \left(x^{2}+y\right)^{17}$

$16 \log x+9 \log y=17 \log \left(x^{2}+y\right)$

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

Differentiating it with respect to $\mathrm{x}$ using the chain rule,

$16 \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})+9 \frac{\mathrm{d}}{\mathrm{dx}}(\log y)=17 \frac{\mathrm{d}}{\mathrm{dx}} \log \left(\mathrm{x}^{2}+\mathrm{y}\right)$

$\frac{16}{x}+\frac{9}{y} \frac{d y}{d x}=17 \cdot \frac{1}{\left(x^{2}+y\right)} \frac{d}{d x}\left(x^{2}+y\right)$

$\frac{16}{x}+\frac{9}{y} \frac{d y}{d x}=\frac{17}{\left(x^{2}+y\right)}\left[2 x+\frac{d y}{d x}\right]$

$\frac{16}{x}+\frac{9}{y} \frac{d y}{d x}=\left[\frac{17}{\left(x^{2}+y\right)}\right] \frac{d y}{d x}+\left[\frac{34 x}{\left(x^{2}+y\right)}\right]$

$\frac{9}{y} \frac{d y}{d x}-\frac{17}{\left(x^{2}+y\right)} \frac{d y}{d x}=\left(\frac{34 x}{x^{2}+y}\right)-\frac{16}{x}$

$\frac{d y}{d x}\left[\frac{9}{y}-\frac{17}{\left(x^{2}+y\right)}\right]=\frac{34 x^{2}-16\left(x^{2}+y\right)}{\left(x^{2}+y\right) x}$

$\frac{d y}{d x}\left[\frac{9\left(x^{2}+y\right)-17 y}{y\left(x^{2}+y\right)}\right]=\frac{34 x^{2}-16 x^{2}-16 y}{\left(x^{2}+y\right) x}$

$\frac{d y}{d x}\left[\frac{9 x^{2}+9 y-17 y}{y\left(x^{2}+y\right)}\right]=\frac{18 x^{2}-16 y}{\left(x^{2}+y\right) x}$

$\frac{d y}{d x}\left[\frac{9 x^{2}+9 y-17 y}{y\left(x^{2}+y\right)}\right]=\frac{2\left(9 x^{2}-8 y\right)}{\left(x^{2}+y\right) x}$

$\frac{d y}{d x}\left[\frac{9 x^{2}-8 y}{y\left(x^{2}+y\right)}\right]=\frac{2\left(9 x^{2}-8 y\right)}{\left(x^{2}+y\right) x}$

$\frac{d y}{d x}=\left[\frac{2\left(9 x^{2}-8 y\right)}{\left(x^{2}+y\right) x}\right]\left[\frac{y\left(x^{2}+y\right)}{9 x^{2}-8 y}\right]$

$\frac{d y}{d x}=\frac{2 y}{x}$

Hence, Proved.