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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 33 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 33 maths textbook solution

Answers (1)

Answer:  \frac{d y}{d x}=\frac{y}{x}

Hint: To differentiate the equation take log on both the sides

Given:  x^{13} y^{7}=(x+y)^{20}

Solution:

         \log \left(x^{13} y^{7}\right)=\log (x+y)^{20}

        \log \left(x^{13}\right)+\log \left(y^{7}\right)=\log (x+y)^{20}

        \frac{d}{d x}(13 \log x+7 \log y)=\frac{d}{d x}(20 \log (x+y))

        13 \cdot \frac{1}{x}+7 \cdot \frac{1}{y} \cdot \frac{d y}{d x}=20 \cdot \frac{1}{x+y} \cdot \frac{d}{d x}(x+y)

        \frac{13}{x}+\frac{7}{y} \frac{d y}{d x}=\frac{20}{x+y} \cdot\left(1+\frac{d y}{d x}\right)

        \left(\frac{7}{y}-\frac{20}{x+y}\right) \cdot \frac{d y}{d x}=\frac{20}{x+y}-\frac{13}{x}

        \frac{7 x+7 y-20 y}{(x+y) y} \frac{d y}{d x}=\frac{20 x-13 x-13 y}{x(x+y)}

        \frac{7 x-13 y}{y(x+y)} \frac{d y}{d x}=\frac{7 x-13 y}{x(x+y)}

        \begin{aligned} &\frac{d y}{d x}=\frac{7 x-13 y}{\sqrt{(x+y)}} \times \frac{y[(x+y)}{\frac{d y}{7 x-13 y}} \\\\ &\frac{d y}{d x}=\frac{y}{x} \end{aligned}

Hence proved

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