# What is the differentiation of 1/log x ?

## Solution :

We have, y = $$1\over log x$$

By using quotient rule in differentiation,

$$dy\over dx$$ = $$log x.{d\over dx}(1) – 1 {d\over dx}(log x)\over (log x)^2$$

$$dy\over dx$$ = $$0 – {1\over x}\over (log x)^2$$ = $$-1\over x (log x)^2$$

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

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