Question

How do you differentiate `f(x) = 1/(x^3-4x)` using the quotient rule?

Answer 1

Quotient rule states that for a function `y=f(x)/g(x)`, `(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/g(x)^2`.

To be honest, I'd better use chain rule, here, but let's do as you asked.

Explanation:

`(df(x))/(dx)=((0)(x^3-4x)-1(3x^2-4))/(x^3-4x)^2`

`(df(x))/(dx)=(cancel((0)(x^3-4x))-1(3x^2-4))/(x^3-4x)^2`

`(df(x))/(dx)=(-1(3x^2-4))/(x^3-4x)^2=(4-3x^2)/(x^3-4x)^2`