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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability3 Marks
Question
Differentiate the function (log x)log x, x > 1, w.r.t to x.
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Answer

Let y = (log x)log x, x > 1
Taking logarithm on both sides
$\Rightarrow$ log y = log (log x)log x = log x $\times$ log (log x)
Differentiating both sides with respect to x, we get
$\frac{1}{\mathrm{y}} \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}[\log \mathrm{x} \times \log (\log \mathrm{x})]$ 
$\Rightarrow$ $\frac{1}{y} \frac{d y}{d x}=\log (\log x) \times \frac{d}{d x}(\log x)+\log x \times \frac{d}{d x}[\log (\log x)]$ 
$\Rightarrow \frac{d y}{d x}=y\left[\log (\log x) \times \frac{1}{x}+\log x \times \frac{1}{\log x} \times \frac{d}{d x}(\log x)\right]$ 
$\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x} \log (\log x)+\frac{1}{x}\right]$ 
$\therefore \frac{d y}{d x}=(\log x)^{\log x}\left[\frac{1}{x}+\frac{\log (\log x)}{x}\right]$

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