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Solution for If X Y + Y X = ( X + Y ) X + Y , Find D Y D X ? - CBSE (Commerce) Class 12 - Mathematics

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Question

If \[x^y + y^x = \left( x + y \right)^{x + y} , \text{ find } \frac{dy}{dx}\] ?

Solution

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

\[ \Rightarrow e^{ \log x^y} + e^{\log y^x } = e^{ \log \left( x + y \right)^\left( x + y \right) } \]

\[ \Rightarrow e^{y \log x} + e^{x \log y} = e^{ \left( x + y \right) \log\left( x + y \right) }\]

Differentiating with respect to x using chain rule and product rule,

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

\[ \Rightarrow e^{y \log x } \left[ y\frac{d}{dx}\left( \log x \right) + \log x\frac{dy}{dx} \right] + e^{x \log y} \left[ x\frac{d}{dx}\log y + \log y\frac{d}{dx}\left( x \right) \right] = e^\left( x + y \right)\log\left( x + y \right) \frac{d}{dx}\left[ \left( x + y \right)\log\left( x + y \right) \right]\]

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

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

\[ \Rightarrow x^y \times \frac{y}{x} + x^y \log x\frac{dy}{dx} + y^x \times \frac{x}{y}\frac{dy}{dx} + y^x \log y = \left( x + y \right)^\left( x + y \right) \left[ 1 \times \left( 1 + \frac{dy}{dx} \right) + \log\left( x + y \right)\left( 1 + \frac{dy}{dx} \right) \right]\]

\[ \Rightarrow x^{y - 1} \times y + x^y \log x\frac{dy}{dx} + y^{x - 1} \times x\frac{dy}{dx} + y^x \log y = \left( x + y \right)^\left( x + y \right) + \left( x + y \right)^\left( x + y \right) \frac{dy}{dx} + \left( x + y \right)^\left( x + y \right) \log\left( x + y \right) + \left( x + y \right)^\left( x + y \right) \log\left( x + y \right)\frac{dy}{dx}\]

\[ \Rightarrow \frac{dy}{dx}\left[ x^y \log x + x y^{x - 1} - \left( x + y \right)^\left( x + y \right) \left\{ 1 + \log\left( x + y \right) \right\} \right] = \left( x + y \right)^\left( x + y \right) \left\{ 1 + \log\left( x + y \right) \right\} - x^{y - 1} \times y - y^x \log y\]

\[ \Rightarrow \frac{dy}{dx} = \left[ \frac{\left( x + y \right)^\left( x + y \right) \left\{ 1 + \log\left( x + y \right) \right\} - y x^{y - 1} - y^x \log y}{x^y \log x + x y^{x - 1} - \left( x + y \right)^\left( x + y \right) \left\{ 1 + \log\left( x + y \right) \right\}} \right]\]

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Solution If X Y + Y X = ( X + Y ) X + Y , Find D Y D X ? Concept: Simple Problems on Applications of Derivatives.
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