Question

If y=${\left(\log x\right)}^x\frac{dy}{dx}$ then $\frac{dy}{dx}$

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

Answer 1

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

$y={\left(\log x\right)}^x.........\left(1\right)$

Take $\log$ both side.

$\log y=\log {\left(\log x\right)}^x$

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

Differentiate both side with respect to $x$

$\frac{d\log y}{dx}=\frac{d\left(x\log \left(\log x\right)\right)}{dx}$

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

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

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

From (1).

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