`y = 1/(ln(x))` Differentiate the function.

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 Differentiate the function.

Rewrite this as: 

`y=(lnx)^-1`

We can take the derivative by power rule and chain rule.  Chain rule is the derivative of the inner function.

`y'= -(ln(x))^-2 * (1/x)`

This simplifies to:  

`y'=-1/(x(lnx)^2)`

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You need to differentiate the function with respect to x, using the quotient rule, such that:

`f'(x) = (1/(ln x))'`

`f'(x) = (1'*(ln x) - 1*(ln x)')/(ln^2 x)`

`f'(x) = (0*ln x - 1/x)/(ln^2 x)`

`f'(x) = -1/(x*ln^2 x)`

Hence, evaluating the derivative of the given function, yields` f'(x) = -1/(x*ln^2 x).`

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