What is the differentiation of 1/log x ?

Solution :

We have, y = \(1\over log x\)

By using quotient rule in differentiation,

\(dy\over dx\) = \(log x.{d\over dx}(1) โ€“ 1 {d\over dx}(log x)\over (log x)^2\)

\(dy\over dx\) = \(0 โ€“ {1\over x}\over (log x)^2\) = \(-1\over x (log x)^2\)

Hence, the differentiation of 1/log x with respect to x is \(-1\over x (log x)^2\).


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