How do you integrate f(x)=(2x)/(x^3-7) using the quotient rule?

1 Answer
Nov 5, 2016

f'(x)=-(4x^3+14)/(x^3-7)^2

Explanation:

The quotient rule states that if f(x)=g(x)/(h(x)), then

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

So, where f(x)=(2x)/(x^3-7), we see that:

f'(x)=(d/dx(2x)*(x^3-7)-2x*d/dx(x^3-7))/(x^3-7)^2

Through the product rules: d/dx(2x)=2 and d/dx(x^3-7)=3x^2:

f'(x)=(2(x^3-7)-2x(3x^2))/(x^3-7)^2

f'(x)=(2x^3-14-6x^3)/(x^3-7)^2

f'(x)=-(4x^3+14)/(x^3-7)^2