Question

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

[1 mark]

• A. $\frac{1}{x}$
• B. $\frac{1}{x}(1+{\log }_{e}10)$
• C. $\frac{1}{x}(1+{\log }_{10}e)$
• D. $1+{\log }_{e}x$

Answer 1

The correct option is C $\frac{1}{x}(1+{\log }_{10}e)$

$y={\log }_{10}e+{\log }_{10}x+{\log }_{e}x$
$\Rightarrow y={\log }_{10}e+\frac{{\log }_{e}x}{{\log }_{e}10}+{\log }_{e}x$
Differentiating w.r.t. $x,$ we get
$\frac{dy}{dx}=0+\frac{1}{{\log }_{e}10}\cdot \frac{1}{x}+\frac{1}{x}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{x}(1+{\log }_{10}e)$