Question

Differentiate the given function w.r.t. $x$:
$(\log x{)}^{\log x}$, $x>1$

• A. $(\log x{)}^{\log x}[\frac{1}{x}-\frac{\log \left(\log x\right)}{x}]$
• B. $(\log x{)}^{\log x}[\frac{1}{x}+\frac{\log \left(\log x\right)}{x}]$
• C. $-(\log x{)}^{\log x}[\frac{1}{x}-\frac{\log \left(\log x\right)}{x}]$
• D. $-(\log x{)}^{\log x}[\frac{1}{x}+\frac{\log \left(2\log x\right)}{x}]$

Answer 1

The correct option is B $(\log x{)}^{\log x}[\frac{1}{x}+\frac{\log \left(\log x\right)}{x}]$

Let $y=(\log x{)}^{\log x}$

Taking logarithm on both the sides, we obtain

$\log y=\log x.\log \left(\log x\right)$

Differentiating both sides with respect to x, we obtain

$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[\log x.\log (\log x\left)\right]$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\log \left(\log x\right).\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left[\log \right(\log x\left)\right]$

$\Rightarrow \frac{dy}{dx}=y\left[\log \right(\log x)\frac{1}{x}+\log x.\frac{1}{\log x}.\frac{d}{dx}(\log x\left)\right]$

$\Rightarrow \frac{dy}{dx}=y\left[\frac{1}{x}\log \right(\log x)+\frac{1}{x}]$

$\therefore \frac{dy}{dx}=(\log x{)}^{\log x}[\frac{1}{x}+\frac{\log \left(\log x\right)}{x}]$