How do you differentiate #f(x) = 1/(x^3-x)# using the quotient rule?

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Answer 1 · 2016-01-11T12:28:47

#f'(x)=(1-3x^2)/(x^3-x)^2#

The quotient rule states that

#d/dx[(g(x))/(h(x))]=(g'(x)h(x)-h'(x)g(x))/[h(x)]^2#

Here, we have

#g(x)=1#
#h(x)=x^3-x#

Differentiate the two functions:

#g'(x)=0#
#h'(x)=3x^2-1#

Plug these in to the quotient rule expression to see that

#f'(x)=(0(x^3-x)-(3x^2-1)(1))/(x^3-x)^2#

Which simplifies to be

#f'(x)=(1-3x^2)/(x^3-x)^2#