Question

Consider the following statements :
Statement 1 : If y = log₁₀x + log_ex then \(\frac{dy}{dx}=\frac{\log_{10}e}{x}+\frac{1}{x}\)
Statement 2 : \(\frac{d}{dx}(\log_{10}x)=\frac{\log x}{\log 10}\ \text{and}\ \frac{d}{dx}(\log_ex)=\frac{\log x}{\log e}\)

• A. Statement 1 is true; statement 2 is false
• B. Statement 1 is false; statement 2 is true
• C. Both statements 1 and 2 are true
• D. Both statements 1 and 2 are false

Answer 1

The Correct Option is A

Statement 1: We are given that \( y = \log_{10} x + \log_e x \). The derivatives of the logarithmic functions are: \[ \frac{d}{dx} (\log_{10} x) = \frac{1}{x \ln 10} \] and \[ \frac{d}{dx} (\log_e x) = \frac{1}{x \ln e} = \frac{1}{x} \] Adding these, we get: \[ \frac{dy}{dx} = \frac{1}{x \ln 10} + \frac{1}{x} = \frac{\log_{10} e}{x} + \frac{1}{x} \] Hence, Statement 1 is true.
Statement 2: The derivatives of \( \log_{10}x \) and \( \log_e x \) are incorrect in the statement. The correct derivatives are: \[ \frac{d}{dx} (\log_{10}x) = \frac{1}{x \ln 10} \] and \[ \frac{d}{dx} (\log_e x) = \frac{1}{x} \] Hence, Statement 2 is false.

The correct answer is (A) : Statement 1 is true; statement 2 is false.