Question

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

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

Answer 1

Answer: (C)

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