Question

If $y={\log }_2{\log }_2\left(x\right)$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{{\log }_{2}e}{{\log }_{e}x}$
• B. $\frac{{\log }_{2}e}{x{\log }_{x}2}$
• C. $\frac{{\log }_{2}x}{{\log }_{e}2}$
• D. $\frac{{\log }_2\mathrm{e}}{{\mathrm{xlog}}_{\mathrm{e}}\mathrm{x}}$

Answer 1

The correct option is D

$\frac{{\log }_2\mathrm{e}}{{\mathrm{xlog}}_{\mathrm{e}}\mathrm{x}}$

Explanation for the correct option:

Step 1: Simplify :

Given that, $y={\log }_2{\log }_2\left(x\right)$.

Using property of logarithmic, ${\log }_{b}a=\frac{\log a}{\log b}$.
$\begin{array}{rcl}y& =& {\log }_2\left[\frac{{\log }_e\left(x\right)}{{\log }_{e}2}\right]\\ & =& \frac{\left({\log }_e\left[{\displaystyle \frac{{\log }_e\left(x\right)}{{\log }_{e}2}}\right]\right)}{{\log }_{e}2}\\ & =& \frac{\left[{\log }_e{\log }_{e}x-{\log }_e{\log }_{e}2\right]}{{\log }_{e}2}\left[\because \log \frac{a}{b}=\mathrm{loga}-\mathrm{logb}\right]\end{array}$

Step 2: Find the $\frac{dy}{dx}$:

Now, differentiate $y$ with respect to $x$we get:
$\begin{array}{rcl}\frac{dy}{dx}& =& \left(\frac{1}{{\log }_{e}2}\right)\left[\left(\frac{1}{{\log }_{e}x}\right)\left(\frac{1}{x}\right)-0\right]\\ & =& \frac{{\log }_{2}e}{x{\log }_{e}x}\left[\because \frac{1}{{\log }_a\left(b\right)}={\log }_b\left(a\right)\right]\end{array}$

Hence, option D is correct.