Question

If $y={\log }_{10}x+{\log }_{x}10+{\log }_{x}x+{\log }_{10}10$, then $\frac{dy}{dx}$

• A. $\frac{1}{x\log 10}-\frac{\log 10}{x(\log x{)}^2}$
• B. $\frac{1}{x{\log }_{2}10}-\frac{1}{x{\log }_{10}\mathrm{e}}$
• C. $\frac{1}{x\log 10}+\frac{\log 10}{x(\log x{)}^2}$
• D. none of these

Answer 1

The correct option is A

$\frac{1}{x\log 10}-\frac{\log 10}{x(\log x{)}^2}$

Explanation for the correct option:

Find the $\frac{dy}{dx}$.

Given that, $y={\log }_{10}x+{\log }_{x}10+{\log }_{x}x+{\log }_{10}10$

Using logarithmic property, ${\log }_{a}b=\frac{\log b}{\log a}$we have:

$\begin{array}{rcl}y& =& \frac{\log x}{\log 10}+\frac{\log 10}{\log x}+\frac{\log x}{\log x}+\frac{\log 10}{\log 10}\\ & =& \log x+\frac{1}{\log x}+1+1\\ & =& \log x+\frac{1}{\log x}+2\end{array}$

Now differentiate $y$ with respect to $x$we get:

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{x}-{\left(\frac{1}{\log x}\right)}^2\frac{1}{x}\\ & =& \frac{1}{x\log 10}-\frac{1}{x{\left(\log x\right)}^2}\end{array}$

Hence, option (A) is correct.