Question

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

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

Answer 1

Answer: (A)

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

Given,
$y={\log }_{10}x+{\log }_{x}10+{\log }_{x}x+{\log }_{10}10$
$\Rightarrow y={\log }_{10}e\cdot {\log }_{e}x+\frac{{\log }_{e}10}{{\log }_{e}x}+1+1$
On differentiating with respect to $x$, we get
$\frac{dy}{dx}=\frac{1}{x}{\log }_{10}e-\frac{{\log }_{e}10}{x{\left({\log }_{e}x\right)}^2}$
$=\frac{1}{x{\log }_{e}10}-\frac{{\log }_{e}10}{x{\left({\log }_{e}x\right)}^2}$