Question:
If \(y = \log_{10} x + \log_e x\), then \(\frac{dy}{dx}\) is equal to
If \(y = \log_{10} x + \log_e x\), then \(\frac{dy}{dx}\) is equal to
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Always convert $\log_a x$ to natural log before differentiation.
Updated On: Apr 30, 2026
- $\frac{1 - \log_{10} e}{x}$
- $\frac{1 + \log_e 10}{x}$
- $x + \log_{10} e$
- $x + \log_e 10$
- $\frac{1}{x}\left[\frac{1}{\log_e 10} + 1\right]$
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The Correct Option is
Solution and Explanation
Concept:
Change of base:
\[
\log_{10} x = \frac{\ln x}{\ln 10}
\]
Step 1: Rewrite function
\[ y = \frac{\ln x}{\ln 10} + \ln x \]
Step 2: Differentiate
\[ \frac{dy}{dx} = \frac{1}{x \ln 10} + \frac{1}{x} \]
Step 3: Take common factor
\[ = \frac{1}{x}\left(\frac{1}{\ln 10} + 1\right) \] Final Conclusion:
Option (E)
Step 1: Rewrite function
\[ y = \frac{\ln x}{\ln 10} + \ln x \]
Step 2: Differentiate
\[ \frac{dy}{dx} = \frac{1}{x \ln 10} + \frac{1}{x} \]
Step 3: Take common factor
\[ = \frac{1}{x}\left(\frac{1}{\ln 10} + 1\right) \] Final Conclusion:
Option (E)
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