What are the first two derivatives of #1/ln(x)#?

1 Answer
vince
Feb 21, 2015

Hello,

Answer If #f(x) = 1/(ln(x))#, then
#f'(x) = - 1/(x \cdot ln(x)^2)# and #f''(x) = (ln(x) + 2)/(x^2 \cdot ln(x)^3)#

First, use the rule #(1/u)' = - (u')/(u^2)# with #u(x) = ln(x)# (therefore #u'(x) = 1/x#). You get #f'(x)#.

Second, use the same rule but with #u(x)=x \cdot ln(x)^2#. Here,
#u'(x) = 1\cdot ln(x)^2 + x \cdot 2\ln(x) \cdot \frac{1}{x} #
#u'(x) = ln(x)^2 + 2\ln(x)#.

Finally, simplify by #ln(x)# to obtain #f''(x)#.

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