# Solve this

Question:

If $x^{13} y^{7}=(x+y)^{20}$, prove that $\frac{d y}{d x}=\frac{y}{x}$

Solution:

Here,

$x^{13} y^{7}=(x+y)^{20}$

Taking log on both sides,

$\log \left(x^{13} y^{7}\right)=\log (x+y)^{20}$

$13 \log x+7 \log y=20 \log (x+y)$

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

Differentiating it with respect to $x$ using the chain rule,

$13 \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})+7 \frac{\mathrm{d}}{\mathrm{dx}}(\log y)=20 \frac{\mathrm{d}}{\mathrm{dx}} \log (\mathrm{x}+\mathrm{y})$

$\frac{13}{\mathrm{x}}+\frac{7}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{20}{\mathrm{x}+\mathrm{y}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\mathrm{y})$

$\frac{7}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{20}{\mathrm{x}+\mathrm{y}}=\frac{20}{\mathrm{x}+\mathrm{y}}-\frac{13}{\mathrm{x}}$

$\frac{d y}{d x}\left[\frac{7}{y}-\frac{20}{x+y}\right]=\frac{20}{x+y}-\frac{13}{x}$

$\frac{d y}{d x}\left[\frac{7(x+y)-20 y}{y(x+y)}\right]=\frac{20 x-13(x+y)}{(x+y) x}$

$\frac{d y}{d x}=\left[\frac{20 x-13(x+y)}{(x+y) x}\right] \times\left[\frac{y(x+y)}{7(x+y)-20 y}\right]$

$\frac{d y}{d x}=\left[\frac{20 x-13 x-13 y}{(x+y) x}\right] \times\left[\frac{y(x+y)}{7 x+7 y-20 y}\right]$

$\frac{d y}{d x}=\frac{y}{x}\left[\frac{7 x-13 y}{7 x-13 y}\right]$

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

Hence, Proved.