Question

If $y = \log_{10} x + \log_x 10 + \log_x x + \log_{10} 10$, then $\frac{dy}{dx} = $

• A. $\frac{1}{x \log_e 10} + \frac{1}{x \log_{10} e}$
• B. $\frac{1}{x \log_e 10} + \frac{\log_e 10}{x (\log_{10} e)^2}$
• C. $\frac{1}{x \log_e 10} - \frac{1}{x \log_{10} e}$
• D. $\frac{1}{x \log_e 10} - \frac{\log_e 10}{x (\log_e x)^2}$

Hint:

Remember that terms like $\log_x x$ and $\log_{k} k$ are just constants equal to 1. Identify them early and drop them from your calculations immediately, since the derivative of any constant is zero!

Answer 1

The Correct Option is D

Step 1: Simplify the constant terms.
In $y=\log_{10}x+\log_x10+\log_x x+\log_{10}10$, note $\log_x x=1$ and $\log_{10}10=1$, which are constants and vanish on differentiation.
Step 2: Convert to natural logs.
$\log_{10}x=\dfrac{\ln x}{\ln10}$ and $\log_x10=\dfrac{\ln10}{\ln x}$, so $y=\dfrac{\ln x}{\ln10}+\dfrac{\ln10}{\ln x}+2$.
Step 3: Differentiate the first term.
$\dfrac{d}{dx}\!\left(\dfrac{\ln x}{\ln10}\right)=\dfrac{1}{\ln10}\cdot\dfrac1x=\dfrac{1}{x\ln10}$.
Step 4: Differentiate the second term.
Writing it as $\ln10\cdot(\ln x)^{-1}$, $\dfrac{d}{dx}=\ln10\cdot(-1)(\ln x)^{-2}\cdot\dfrac1x=-\dfrac{\ln10}{x(\ln x)^2}$.
Step 5: Add the pieces.
\[ \frac{dy}{dx}=\frac{1}{x\ln10}-\frac{\ln10}{x(\ln x)^2}. \]
Step 6: Match the key notation.
With $\ln=\log_e$, this is $\dfrac{1}{x\log_e10}-\dfrac{\log_e10}{x(\log_e x)^2}$, option (D). \[ \boxed{\dfrac{1}{x\log_e10}-\dfrac{\log_e10}{x(\log_e x)^2}} \]