A) e
B) 1
C) 0
D) \[{{\log }_{e}}x\,\,{{e}^{{{\log }_{e}}ex}}\]
Correct Answer: A
Solution :
\[y={{e}^{1+{{\log }_{e}}x}}={{e}^{1}}.{{e}^{{{\log }_{e}}x}}=e.x\Rightarrow \frac{dy}{dx}=e\].
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