How do you differentiate #y= ln( 1/x)#?

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Answer 1 · 2016-07-31T20:44:40

#= - 1/x#

doing it the long way:

#d/dx ln (f(x)) = 1/(f(x)) * f'(x)#

here #f(x) = 1/x = x^(-1)# so by the power rule #d/dx (x^(-1)) = - x^(-2) = -1/x^2#

so

#d/dx ln (1/x) = 1/(1/x) * -1/x^2 = - 1/x#

speeding up a little:

# ln (1/x) = ln x^(-1) = - ln x#

#d/dx (- ln x) = - d/dx (ln x) = - 1/x (x)' = - 1/x#