Question

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

Answer 1

`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`