What is the second derivative of #x/(x-1)# and the first derivative of #2/x#?

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Answer 1 · 2015-04-07T01:46:28

Question 1
If #f(x) = (g(x))/(h(x))# then by the Quotient Rule

#f'(x) = (g'(x)*h(x) - g(x)*h'(x))/((g(x))^2)#

So if #f(x) = x/(x-1)#
then the first derivative
#f'(x) = ((1)(x-1) - (x)(1))/x^2#

#= - 1/x^2 = - x^(-2)#

and the second derivative is
#f''(x) = 2x^-3#

Question 2
If #f(x) = 2/x# this can be re-written as

#f(x) = 2x^-1#

and using standard procedures for taking the derivative
#f'(x) = -2x^-2#
or, if you prefer
#f'(x) = - 2/x^2#