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

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

Answers (1)

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

Solution:  

        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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