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  • Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 38

Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 38

Answers (1)

Answer:  \frac{d y}{d x}=\frac{x(1+\log (x+y)-y x \log y}{y \log x+x-y(1+\log x+y)}

Hint:  To solve this we add log on both sides

Given:   x^{y}+y^{x}=(x+y)^{x+y}

Solution:  

Taking log on both sides,

        \begin{aligned} &\log x^{y}+\log y^{x}=\log (x+y)^{x+y} \\\\ &y \log x+x \log y=(x+y)+\log (x+y) \end{aligned}                \left[\because \frac{d}{d x}(u-v)=u . d v+v \cdot d u\right]

        y \cdot \log x+x \log y^{\prime}+\frac{x}{y} y^{\prime}+\log y=\frac{(x+y)}{(x+y)}\left(1+y^{\prime}\right)+\log (x+y)\left(1+y^{\prime}\right)

        \begin{aligned} &=\left(1+y^{\prime}\right)(1+\log (x+y)) \\\\ &\frac{y}{x}+\log y+y^{\prime}\left(\log x+\frac{x}{y}\right)=1+\log (x+y)+y^{\prime}(1+\log (x+y)) \end{aligned}

        y^{\prime}\left(\log x+\frac{x}{y}-(1+\log (x+y))=1+\log (x+y)-\left(\frac{y}{x}+\log y\right)\right.

        y^{\prime}=\frac{x\left(1+\log (x+y)^{\prime-y}+x \log y\right)}{y \log x+x-y(1+\log (x+y))}

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