Question

If $y={\log }_{a}x+{\log }_{x}a+{\log }_{x}x+{\log }_{a}a$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

The correct option is D

$\frac{1}{x\log a}-\frac{\left(\log a\right)}{x(\log x{)}^2}$

Explanation for the correct option.

Step 1: Simplify

Given that, $y={\log }_{a}x+{\log }_{x}a+{\log }_{x}x+{\log }_{a}a$

Using the property of logarithmic, ${\log }_{a}x=\frac{\log x}{\log a}$ we have:
$\begin{array}{rcl}y& =& \frac{\log x}{\log a}+\frac{\log a}{\log x}+\frac{\log x}{\log x}+\frac{\log a}{\log a}\\ & =& \frac{\log x}{\log a}+\frac{\log a}{\log x}+1+1\\ & =& \frac{\log x}{\log a}+\frac{\log a}{\log x}+2\end{array}$

Step 2: Find the $\frac{dy}{dx}$
Now differentiate $y$ with respect to $x$.
$\begin{array}{rcl}\frac{\mathrm{dy}}{dx}& =& \frac{1}{\mathrm{xloga}}-\left[\frac{\left(\log a\right)}{(\log x{)}^2}\right]\frac{1}{x}\left[\frac{d\left(\log x\right)}{dx}=\frac{1}{x}\right]\\ & =& \frac{1}{x\log a}-\frac{\left(\log a\right)}{x(\log x{)}^2}\end{array}$
Hence, option D is correct.