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# Solution for If X 13 Y 7 = ( X + Y ) 20 Prove that D Y D X = Y X ? - CBSE (Commerce) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

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

#### Solution

$\text{ We have,} x^{13} y^7 = \left( x + y \right)^{20}$

Taking log on both sides,

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

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

Differentiating with respect to x using chain rule,

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

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

$\Rightarrow \frac{13}{x} + \frac{7}{y}\frac{dy}{dx} = \frac{20}{x + y}\left[ 1 + \frac{dy}{dx} \right]$

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

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

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

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

$\Rightarrow \frac{dy}{dx} = \frac{y}{x}\left( \frac{7x - 13y}{7x - 13y} \right)$

$\Rightarrow \frac{dy}{dx} = \frac{y}{x}$

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Solution for question: If X 13 Y 7 = ( X + Y ) 20 Prove that D Y D X = Y X ? concept: Simple Problems on Applications of Derivatives. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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