Question

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

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

Answer 1

The correct option is A $\frac{1}{x{\log }_{e}10}-\frac{{\log }_{e}10}{x({\log }_{e}x{)}^2}$

$\frac{d}{dx}\left({\log }_{10}x+{\log }_{x}10+{\log }_{x}x+{\log }_{10}10\right)$

$=\frac{1}{x{\log }_{e}10}+\frac{(-{\log }_{e}10\left(\frac{1}{x}\right))}{{\left({\log }_{e}x\right)}^2}$

$=\frac{1}{x{\log }_{e}10}-\frac{{\log }_{e}10}{x{\left({\log }_{e}x\right)}^2}$