Question:

Consider the following statements :
Statement 1 : If y = log10x + logex then dydx=log10ex+1x\frac{dy}{dx}=\frac{\log_{10}e}{x}+\frac{1}{x}
Statement 2 : ddx(log10x)=logxlog10 and ddx(logex)=logxloge\frac{d}{dx}(\log_{10}x)=\frac{\log x}{\log 10}\ \text{and}\ \frac{d}{dx}(\log_ex)=\frac{\log x}{\log e}

Updated On: Apr 1, 2025
  • Statement 1 is true; statement 2 is false
  • Statement 1 is false; statement 2 is true
  • Both statements 1 and 2 are true
  • Both statements 1 and 2 are false
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The Correct Option is A

Solution and Explanation

Statement 1: We are given that y=log10x+logex y = \log_{10} x + \log_e x . The derivatives of the logarithmic functions are: ddx(log10x)=1xln10 \frac{d}{dx} (\log_{10} x) = \frac{1}{x \ln 10} and ddx(logex)=1xlne=1x \frac{d}{dx} (\log_e x) = \frac{1}{x \ln e} = \frac{1}{x} Adding these, we get: dydx=1xln10+1x=log10ex+1x \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 log10x \log_{10}x and logex \log_e x are incorrect in the statement. The correct derivatives are: ddx(log10x)=1xln10 \frac{d}{dx} (\log_{10}x) = \frac{1}{x \ln 10} and ddx(logex)=1x \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.

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