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  4. The value of (d/dx)[xn log a xex] is

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Q. The value of $ \frac{d}{dx}[{{x}^{n}}\,{{\log }_{a}}\,x{{e}^{x}}] $ is

J & K CETJ & K CET 2013Continuity and Differentiability

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\} $