A) \[{{e}^{x}}.{{x}^{2}}(x+3)\]
B) \[{{e}^{x}}.x(x+3)\]
C) \[{{e}^{x}}+\frac{3}{x}\]
D) None of these
Correct Answer: A
Solution :
\[{{e}^{x+3\log x}}={{e}^{x}}.{{e}^{3\log x}}={{e}^{x}}.{{e}^{\log {{x}^{3}}}}={{e}^{x}}.{{x}^{3}}\] Therefore\[y={{e}^{x}}.{{x}^{3}}\Rightarrow \frac{dy}{dx}={{e}^{x}}.3{{x}^{2}}+{{x}^{3}}.{{e}^{x}}={{e}^{x}}{{x}^{2}}(3+x)\]
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