#### Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 59

Answer:  $\frac{d y}{d x}=\frac{x-y}{x \log x}$

Hint: To solve this equation we use chain rule and quotient rule

Given: $x=e^{\frac{x}{y}}$

Solution:  we have $x=e^{\frac{x}{y}}$

Diff w.r.t x we get                            $\left[\because \frac{d}{d x} e^{x}=e^{x}\right]$

$1=e^{\frac{x}{y}} \frac{d}{d x}\left(\frac{x}{y}\right)$                        [Chain rule]

$1=e^{\frac{x}{y}} \times \frac{y \frac{d}{d x} x-x \frac{d}{d x} y}{y^{2}} \quad\left[\because\left(\frac{u}{v}\right)^{\prime}=\frac{u r v-v r u}{v^{2}}\right]$

\begin{aligned} &y^{2}=e^{\frac{x}{y}}\left(y-x \frac{d y}{d x}\right) \\\\ &y^{2}=y \cdot e^{\frac{x}{y}}-x \frac{d y}{d x} e^{\frac{x}{y}} \end{aligned}

\begin{aligned} &x \frac{d y}{d x} e^{\frac{x}{y}}=y \cdot e^{\frac{x}{y}}-y^{2} \\\\ &\frac{d y}{d x}=\frac{y\left(e^{\frac{x}{y}}-y\right)}{x \cdot e^{\frac{x}{y}}} \end{aligned}

$x. \frac{d y}{d x}=\frac{1}{\log x}(x-y) \quad\left[\because x=e^{\frac{x}{y}}, \log x=\frac{x}{y}\right]$

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