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If X Y Log ( X + Y ) = 1 ,Prove that D Y D X = − Y ( X 2 Y + X + Y ) X ( X Y 2 + X + Y ) ? - Mathematics

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Question

If \[xy \log \left( x + y \right) = 1\] ,Prove that \[\frac{dy}{dx} = - \frac{y \left( x^2 y + x + y \right)}{x \left( x y^2 + x + y \right)}\] ?

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Solution

\[\text{ We have, xy }  \log\left( x + y \right) = 1\]

Differentiating it with respect to x,

\[\Rightarrow \frac{d}{dx}\left[ xy \log\left( x + y \right) \right] = \frac{d}{dx}\left( 1 \right)\]

\[ \Rightarrow xy\frac{d}{dx}\log\left( x + y \right) + x \log\left( x + y \right)\frac{d y}{d x} + y \log\left( x + y \right)\frac{d}{dx}\left( x \right) = 0 \left[ \text{ using chain rule and product rule }  \right]\]

\[ \Rightarrow xy\left( \frac{1}{x + y} \right)\frac{d}{dx}\left( x + y \right) + x \log\left( x + y \right)\frac{d y}{d x} + y \log\left( x + y \right)\left( 1 \right) = 0\]

\[ \Rightarrow \left( \frac{xy}{x + y} \right) \left( 1 + \frac{d y}{d x} \right) + x \log\left( x + y \right)\frac{d y}{d x} + y \log\left( x + y \right) = 0\]

\[ \Rightarrow \left( \frac{xy}{x + y} \right)\frac{d y}{d x} + \left( \frac{xy}{x + y} \right) + x\left( \frac{1}{xy} \right)\frac{d y}{d x} + y\left( \frac{1}{xy} \right) = 0 \left[ \because xy \log\left( x + y \right) = 1 \right]\]

\[ \Rightarrow \frac{d y}{d x}\left[ \frac{xy}{x + y} + \frac{1}{y} \right] = - \left[ \frac{1}{x} + \frac{xy}{x + y} \right]\]

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

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

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

Hence proved

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Chapter 11: Differentiation - Exercise 11.04 [Page 75]

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RD Sharma Mathematics [English] Class 12
Chapter 11 Differentiation
Exercise 11.04 | Q 20 | Page 75

Video TutorialsVIEW ALL [1]

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