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

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