Solution : We have y = \(log x^2\) By using chain rule in differentiation, let u = \(x^2\) \(\implies\)ย \(du\over dx\) = 2x And, y = log u \(\implies\) \(dy\over du\) = \(1\over u\) = \(1\over x^2\) Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\) \(\implies\) \(dy\over dx\) = \(1\over u\).\(du\over dx\) \(\implies\) \(dy\over โฆ
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 โฆ
Solution : We have, y = log ( log x ) By using chain rule in differentiation, \(dy\over dx\) = \(1\over log x\).\(1\over x\) \(\implies\) \(dy\over dx\) = \(1\over x log x\) Hence, the differentiation of log log x with respect to x is \(1\over x log x\). Similar Questions What is the differentiation of โฆ
