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Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 44 maths

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Answer: \frac{d y}{d x}=\frac{(\log y)^{2}}{\log y-1}

Hint: To solve this equation we convert terms in log

Given: e^{y}=y^{x}



        y \log e=x \log y \quad\left[\because \log a^{b}=b \log a\right]

        y=x \log y                    ...................(1)

        \begin{aligned} &\frac{d y}{d x}=x \cdot \frac{1}{y} \frac{d y}{d x}+\log y \\\\ &\frac{d y}{d x}-\frac{x}{y} \frac{d y}{d x}=\log y \end{aligned}

        \begin{aligned} &\frac{d y}{d x}\left(1-\frac{x}{y}\right)=\log y \\\\ &\frac{d y}{d x}=\frac{\log y}{\left(1-\frac{x}{y}\right)} \end{aligned}

        \begin{aligned} &\frac{d y}{d x}=\frac{\log y}{\left(1-\frac{1}{\log y}\right)} \\\\ &\frac{d y}{d x}=\frac{(\log y)^{2}}{(\log y-1)} \end{aligned}

Hence proved

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