{"id":7865,"date":"2021-11-09T18:46:32","date_gmt":"2021-11-09T13:16:32","guid":{"rendered":"https:\/\/mathemerize.com\/?p=7865"},"modified":"2021-11-09T19:29:10","modified_gmt":"2021-11-09T13:59:10","slug":"what-is-the-differentiation-of-x-log-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-the-differentiation-of-x-log-x\/","title":{"rendered":"What is the Differentiation of x log x ?"},"content":{"rendered":"

Solution :<\/h2>\n

We have, y = x log x<\/p>\n

By using product rule in differentiation<\/a>,<\/p>\n

\\(dy\\over dx\\) = log x \\(dy\\over dx\\)(x) + x \\(dy\\over dx\\) (log x)<\/p>\n

\\(dy\\over dx\\) = log x + x\/x<\/p>\n

\\(\\implies\\) \\(dy\\over dx\\) = log x + 1<\/p>\n

Hence, the differentiation of x log x with respect to x is log x + 1.<\/p>\n

\n

Similar Questions<\/h3>\n

What is the differentiation of 1\/log x ?<\/a><\/p>\n

What is the differentiation of \\(log x^2\\) ?<\/a><\/p>\n

What is the differentiation of log sin x ?<\/a><\/p>\n

What is the Differentiation of log log x ?<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

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 …<\/p>\n

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