MCQ
If $y = {\log _{10}}{x^2}$, then ${{dy} \over {dx}}$ is equal to
- A${2 \over x}$
- ✓${2 \over {x{{\log }_e}10}}$
- C${1 \over {x{{\log }_e}10}}$
- D${1 \over {10x}}$
$y = \frac{{{{\log }_e}{x^2}}}{{{{\log }_e}10}}$, $\left( \because {{\log }_{a}}b=\frac{{{\log }_{e}}b}{{{\log }_{e}}a} \right)$
$y = \frac{{2{{\log }_e}x}}{{{{\log }_e}10}}$,
$\therefore \frac{{dy}}{{dx}} = \frac{2}{{x{{\log }_e}10}}$.
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