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If $y=x^{n} \log x+x(\log x)^{n}$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]$
Given, $y=x^{n} \log x+x(\log x)^{n}$
$$
\begin{array}{c}
\frac{d y}{d x}=n x^{n-1} \log x+x^{n} \cdot \frac{1}{x}+x n(\log x)^{n-1}\left(\frac{1}{x}\right) \\
+1 \cdot(\log x)^{n} \\
=x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]
\end{array}
$$
$$
\begin{array}{c}
\frac{d y}{d x}=n x^{n-1} \log x+x^{n} \cdot \frac{1}{x}+x n(\log x)^{n-1}\left(\frac{1}{x}\right) \\
+1 \cdot(\log x)^{n} \\
=x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]
\end{array}
$$
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