Let \(u = \log_{10}x = \frac{\ln x}{\ln 10}\) and \(v = \log_x 10 = \frac{\ln 10}{\ln x}\).

\(\frac{du}{dx} = \frac{1}{x\ln 10}\) and \(\frac{dv}{dx} = -\frac{\ln 10}{x(\ln x)^2}\).

\(\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/(x\ln 10)}{-\ln 10/(x(\ln x)^2)} = -\frac{(\ln x)^2}{(\ln 10)^2}\)

(Or directly: \(u = \frac{1}{v}\), so \(\frac{du}{dv} = -\frac{1}{v^2} = -u^2\).)

Answer: (A)