Question

The derivative of $Y = \log_{10}^{x} + \log_{x}^{10} + \log_{x}^{8} + \log_{10}^{10}$

• A. $\frac{1}{x \log 10} - \frac{\log 10}{x (\log x)^2}$
• B. $\frac{\log 10}{(x \log x)^2}$
• C. $\log x - \log 10$
• D. $\frac{1}{x \log 10} - \frac{1}{x \log x}$

Answer 1

Answer: (A)

$\frac{1}{x \log 10} - \frac{\log 10}{x (\log x)^2}$

Answer 2

The derivative of $Y = \log_{10} x + \log_x 10 + \log_{10} 8 + \log_{10} 10$ is calculated.

Using the change of base formula for logarithms, $\log_b a = \frac{\ln a}{\ln b}$: $Y = \frac{\ln x}{\ln 10} + \frac{\ln 10}{\ln x} + \frac{\ln 8}{\ln 10} + \frac{\ln 10}{\ln 10}$ $Y = \frac{1}{\ln 10} \ln x + \ln 10 (\ln x)^{-1} + \frac{\ln 8}{\ln 10} + 1$

Differentiating term by term with respect to $x$:

1. $\frac{d}{dx} \left( \frac{1}{\ln 10} \ln x \right) = \frac{1}{\ln 10} \cdot \frac{1}{x} = \frac{1}{x \ln 10}$
2. $\frac{d}{dx} \left( \ln 10 (\ln x)^{-1} \right) = \ln 10 \cdot (-1) (\ln x)^{-2} \cdot \frac{1}{x} = -\frac{\ln 10}{x (\ln x)^2}$
3. $\frac{d}{dx} \left( \frac{\ln 8}{\ln 10} \right) = 0$ (since it's a constant)
4. $\frac{d}{dx} (1) = 0$ (since it's a constant)

Combining these terms, the derivative is: $\frac{dY}{dx} = \frac{1}{x \ln 10} - \frac{\ln 10}{x (\ln x)^2}$

Assuming that log in the options refers to the natural logarithm (ln), this result exactly matches the first option. This suggests that the original question notation $\log_{10}^{x}$ was intended as $\log_{10} x$, and $\log_{x}^{8}$ and $\log_{10}^{10}$ were intended as constants $\log_{10} 8$ and $\log_{10} 10$ respectively, or that the terms with base 'x' were meant to be constants. The interpretation that yields a matching option is: $Y = \log_{10} x + \log_x 10 + (\text{constants})$