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Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 49 maths textbook solution

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Answer:  \frac{d y}{d x}=\frac{-y\left(x^{2} y+x+y\right)}{x\left(x y^{2}+x+y\right)}

Hint: To solve this equation we solve the log term firstly

Given: x y \log (x+y)=1


        x y \log (x+y)=1

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

        \begin{aligned} &\frac{d}{d x} \log (x+y)=\frac{d}{d x}(1) \\\\ &y \cdot \log (x+y)+\frac{d y}{d x} \cdot x \cdot \log (x+y)+x \cdot y \cdot \frac{1}{x+y}\left(1+\frac{d y}{d x}\right)=0 \end{aligned}

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

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

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

Hence proved

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