Question

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

• A. $\frac{y}{y-1}$
• B. $\frac{y}{x}$
• C. $\frac{log_{10}\,e}{x}\left(\frac{y}{y-1}\right)$
• D. None of these

Answer 1

Answer: (C)

We have, $y = log_{10}\, x + log_{e}y$
$\therefore \frac{dy}{dx}=\frac{1}{x}log_{10}\,e+\frac{1}{y} \frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{log_{10}\,e}{x}\left(\frac{y}{y-1}\right)$