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  • Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 31

Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 31

Answers (1)

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

Hint: Differentiate the equation taking log on both sides

Given:  y=x^{x}+x^{\frac{1}{x}}

Solution:  y=x^{x}+x^{\frac{1}{x}}

            \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}

Let u=x^{x}

        \begin{aligned} &\log u=\log x^{x} \\\\ &\frac{d}{d x}(\log u)=\log x^{x} \\\\ &\frac{d}{d x}(\log u)=\frac{d}{d x}(x \log x) \end{aligned}

        \begin{aligned} &\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d x}{d x} \cdot \log x+x \frac{d}{d x}(\log x) \\\\ &\left.\frac{d u}{d x}=u[\log x+x] \cdot \frac{1}{x}\right] \\\\ &\frac{d u}{d x}=x^{x}[\log x+1] \end{aligned}

Let v=x^{\frac{1}{x}}

        \log v=\log x^{\frac{1}{x}}

        \begin{aligned} &\frac{d}{d x}(\log v)=\frac{d}{d x}\left(\frac{1}{x} \log x\right) \\\\ &\frac{1}{v} \cdot \frac{d v}{d x}=\frac{1}{x} \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}\left(\frac{1}{x}\right) \end{aligned}

        \begin{aligned} &\frac{d v}{d x}=v\left[\frac{1}{x^{2}}+\log x \cdot\left(\frac{-1}{x^{2}}\right)\right] \\\\ &=x^{\frac{1}{x}}\left[\frac{1-\log x}{x^{2}}\right] \end{aligned}

        \begin{aligned} &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \\\\ &=x^{x}(\log x+1)+x^{\frac{1}{x-2}}(1-\log x) \end{aligned}

 

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