Question

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

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Solution

$\mathrm{We}\mathrm{have},{x}^{13}{y}^{7}={\left(x+y\right)}^{20}\phantom{\rule{0ex}{0ex}}$ Taking log on both sides, $\mathrm{log}\left({x}^{13}{y}^{7}\right)=\mathrm{log}{\left(x+y\right)}^{20}\phantom{\rule{0ex}{0ex}}⇒13\mathrm{log}x+7\mathrm{log}y=20\mathrm{log}\left(x+y\right)$ Differentiating with respect to x using chain rule, $13\frac{d}{dx}\left(\mathrm{log}x\right)+7\frac{d}{dx}\left(\mathrm{log}y\right)=20\frac{d}{dx}\mathrm{log}\left(x+y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{13}{x}+\frac{7}{y}\frac{dy}{dx}=\frac{20}{x+y}\frac{d}{dx}\left(x+y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{13}{x}+\frac{7}{y}\frac{dy}{dx}=\frac{20}{x+y}\left[1+\frac{dy}{dx}\right]\phantom{\rule{0ex}{0ex}}⇒\frac{7}{y}\frac{dy}{dx}-\frac{20}{x+y}\frac{dy}{dx}=\frac{20}{x+y}-\frac{13}{x}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}\left[\frac{7}{y}-\frac{20}{x+y}\right]=\frac{20}{x+y}-\frac{13}{x}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}\left[\frac{7\left(x+y\right)-20y}{y\left(x+y\right)}\right]=\left[\frac{20x-13\left(x+y\right)}{x\left(x+y\right)}\right]\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\left[\frac{20x-13x-13y}{x\left(x+y\right)}\right]\left[\frac{y\left(x+y\right)}{7x+7y-20y}\right]\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{y}{x}\left(\frac{7x-13y}{7x-13y}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{y}{x}$

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