First slide

Derivatives

Question

$If y = (\log x)^{x} then \frac{d y}{d x} =$

Moderate

A

$(\log x)^{x} [\log (\log x) + \frac{1}{\log x}]$
B

$(\log x)^{x} [\log (\log x) - \frac{1}{\log x}]$
C

$- (\log x)^{x} [\log (\log x) + \frac{1}{\log x}]$
D

$- (\log x)^{x} [\log (\log x) - \frac{1}{\log x}]$

Solution

$y = (\log x)^{x}$
Take logarithm on both sides
$\log y = \log (\log x)^{x}$
Differentiate both sides
$\begin{matrix} \frac{1}{y} \frac{d y}{d x} = \log (\log x) + x \frac{1}{\log x} \frac{d}{d x} \log x \\ \frac{1}{y} \frac{d y}{d x} = \log (\log x) + x \frac{1}{\log x} \frac{d}{d x} \log x \\ = \log (\log x) + \frac{x}{\log x} \frac{1}{x} \end{matrix}$
\ $Therefore, \frac{d y}{d x} = y [\log (\log x) + \frac{1}{\log x}] = (\log x)^{x} [\log (\log x) + \frac{1}{\log x}]$

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