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#### Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 1 subquestion (ix) maths textbook solution

$\frac{-(1+log\: x)}{x^{2}(log\: x)^{2}}$

Hint:

You must know about derivative of log(log x)

Given:

$log(log\: x)$

Solution:

$Let\: \:y= log(log\: x)$

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

Use Quotient rule

\begin{aligned} &\text { As } \frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}}\\ &\text { Where } u=1 \& v=x \log x\\ &\text { Use multiplicative rule }\\ &\text { As } U V=U V^{1}+U^{1} V\\ &\text { Where } U=1 \& V=\log x \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{\frac{d}{d x} 1 x \log x-1 \frac{d}{d x} x \log x}{(x \log x)^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{0 . x \log x-x \log x}{x^{2}(\log x)^{2}} \quad\left(\frac{d}{d x} \log x=\frac{1}{x}, \frac{d}{d x} 1=0, \frac{d}{d x} x=1\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{-\log x-1}{x^{2}(\log x)^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-(1+\log x)}{x^{2}(\log x)^{2}} \end{aligned}

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