Question

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

• 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. ${{x}^{n}}\,{{\log }_{a}}x.{{e}^{x}}$

Answer 1

Answer: (B)

{{e}^{x}}{{x}^{n-1}}\,\left\\{ x{{\log }_{a}}\,x+\frac{1}{{{\log }_{e}}\,a}+n\,{{\log }_{a}}x \right\\}

Answer 2

$\frac{d}{dx}[{{x}^{n}}.{{\log }_{a}}x.{{e}^{x}}]$
$={{x}^{n}}.\frac{d}{dx}\\{{{\log }_{a}}\,x.{{e}^{x}}\\}+{{\log }_{a}}\,x.{{e}^{x}}\frac{d}{dx}({{x}^{n}})$
={{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\\}