How do you differentiate #f(x) = (x)/(x^4-x^3+6)# using the quotient rule?

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Answer 1 · 2016-02-07T15:11:53

#f'(x)=(-3x^4+2x^3+6)/(x^4-x^3+6)^2#

The quotient rule states that

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

Applying this to the given function, we see that

#f'(x)=((x^4-x^3+6)d/dx[x]-xd/dx[x^4-x^3+6])/(x^4-x^3+6)^2#

Both of these derivatives can be found through the power rule:

#f'(x)=((x^4-x^3+6)(1)-(x)(4x^3-3x^2))/(x^4-x^3+6)^2#

Simplified, this gives

#f'(x)=(x^4-x^3+6-4x^4+3x^3)/(x^4-x^3+6)^2#

#f'(x)=(-3x^4+2x^3+6)/(x^4-x^3+6)^2#