A) \[{{e}^{x}}\,{{\log }_{a}}\,x+\frac{{{x}^{n-1}}}{{{\log }_{e}}\,a}\]
B) \[{{e}^{x}}{{x}^{n-1}}\,\left\{ x{{\log }_{a}}\,x+\frac{1}{{{\log }_{e}}\,a}+n\,{{\log }_{a}}x \right\}\]
C) \[n{{x}^{n-1}}\,{{\log }_{a}}\,x{{e}^{x}}\]
D) \[{{x}^{n}}\,{{\log }_{a}}x.{{e}^{x}}\]
Correct Answer: B
Solution :
\[\frac{d}{dx}[{{x}^{n}}.{{\log }_{a}}x.{{e}^{x}}]\] \[={{x}^{n}}.\frac{d}{dx}\{{{\log }_{a}}\,x.{{e}^{x}}\}+{{\log }_{a}}\,x.{{e}^{x}}\frac{d}{dx}({{x}^{n}})\] \[={{x}^{n}}\left\{ {{\log }_{a}}x.\frac{d}{dx}\,{{e}^{x}}+{{e}^{x}}\frac{d}{dx}{{\log }_{a}}x \right\}\] \[+{{e}^{x}}\,{{\log }_{a}}\,x.n{{x}^{n-1}}\] \[={{x}^{n}}\left\{ {{e}^{x}}\,{{\log }_{a}}\,x+\frac{{{e}^{x}}}{x}.\frac{1}{{{\log }_{e}}a} \right\}\] \[+n{{x}^{n-1}}.{{e}^{x}}{{\log }_{a}}\,x\] \[={{x}^{n-1}}\,{{e}^{x}}\left\{ x\,{{\log }_{a}}x+\frac{1}{{{\log }_{e}}\,a} \right\}+n{{x}^{n-1}}.{{e}^{x}}\,{{\log }_{a}}x\] \[={{x}^{n-1}}\,.\,{{e}^{x}}\left\{ x\,{{\log }_{a}}x+\frac{1}{{{\log }_{e}}\,a}+n\,{{\log }_{a}}x \right\}\]
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