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  • Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 3

Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 3

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Answer:   x(\log x)^{x-1}\{1+\log x \cdot \log (\log x)\}

Hint:  \text { Let } u=(\log x)^{x}, v=\log x

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

Given:  (\log x)^{x} \text { w.r.t } \log x

Explanation:

\text { Let } u=(\log x)^{x}, v=\log x

u=(\log x)^{x}

Taking log both sides,

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

Differentiate both sides w.r.t. x

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

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

\begin{aligned} &v=\log x \\\\ &\frac{d v}{d x}=\frac{1}{x} \end{aligned}

\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{(\log x)^{x-1}[1+\log x \cdot \log (\log x)]}{\frac{1}{x}}

      =x(\log x)^{x-1}\{1+\log x \cdot \log (\log x)\}

 

Posted by

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