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

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

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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)}

 

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