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Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 37 maths textbook solution

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Answer:  \frac{d y}{d x}=\frac{-(y(y+x \log y))}{x(y \log x+x)}

Hint:  Adding log on both sides of equation

Given: -x^{y} \cdot y^{x}=1


Taking log on both sides,

        \begin{aligned} &\log x^{y}+\log y^{x}=0 \\\\ &\mathrm{y} \log x+x \log y=0 \end{aligned}

        \log x \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log x)+\log y \frac{d}{d x} x+x \frac{d}{d x}(\log y)=0

        \begin{aligned} &\log x \frac{d y}{d x}+\frac{y}{x}+\log y+\frac{x}{y} \frac{d y}{d x}=0 \\\\ &\frac{d y}{d x}\left(\log x+\frac{x}{y}\right)=-\left(\frac{y}{x}+\log y\right) \end{aligned}

        \frac{d y}{d x}=\frac{-\left(\frac{y}{x}+\log y\right)}{\left(\log x+\cdot \frac{x}{y}\right)} \frac{d y}{d x}

        \begin{aligned} &\frac{d y}{d x}=\frac{-(y(y+x \log y)}{x(y \log x+x)} \\\\ &\frac{d y}{d x}=\frac{-(y(y+x \log y)}{x(y \log x+x)} \end{aligned}

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