A) \[2{{x}^{2}}\,{{\log }_{e}}\,10\]
B) \[\frac{{{\log }_{10}}\,e}{2{{x}^{2}}}\]
C) \[\frac{{{\log }_{e}}\,10}{2{{x}^{2}}}\]
D) \[{{x}^{2}}\,{{\log }_{e}}\,10\]
Correct Answer: B
Solution :
Let \[u={{\log }_{10}}x=\frac{{{\log }_{e}}x}{{{\log }_{e}}10}={{\log }_{10}}\,e\,{{\log }_{e}}x\] \[\therefore \] \[\frac{du}{dx}=\frac{{{\log }_{10}}\,e}{x}\] and \[v={{x}^{2}}\] \[\therefore \] \[\frac{dv}{dx}=2x\] Now, \[\frac{du}{dv}=\frac{{{\log }_{10}}e}{x+2x}=\frac{{{\log }_{10}}e}{2{{x}^{2}}}\]
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