Q93 / 150 · MHT CET 2008 · Mathematics

If $y=x^{n} \log x+x(\log x)^{n}$, then $\frac{d y}{d x}$ is equal to

  • Correct answer
    A
    $x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]$
  • B
    $x^{n-2}(1+n \log x)+(\log x)^{n-1}[n+\log x]$
  • C
    $x^{n-1}(1+n \log x)+(\log x)^{n-1}[n-\log x]$
  • D
    None of the above

Solution

Given, $y=x^{n} \log x+x(\log x)^{n}$ $$ \begin{array}{c} \frac{d y}{d x}=n x^{n-1} \log x+x^{n} \cdot \frac{1}{x}+x n(\log x)^{n-1}\left(\frac{1}{x}\right) \\ +1 \cdot(\log x)^{n} \\ =x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x] \end{array} $$

Hence, the correct answer is option: 'A'.

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