Solution :
We have, y = x log x
By using product rule in differentiation,
\(dy\over dx\) = log x \(dy\over dx\)(x) + x \(dy\over dx\) (log x)
\(dy\over dx\) = log x + x/x
\(\implies\) \(dy\over dx\) = log x + 1
Hence, the differentiation of x log x with respect to x is log x + 1.
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