Question

How do you find the derivative of `F(x)= (2x^3-1)/x^2`?

Answer 1

The quotient rule states that given a function `F(x) = (g(x))/(h(x)), (dF)/dx = ((dg)/(dx) * h(x) - g(x) * (dh)/dx)/(h(x))^2`. See explanation for solution.

Explanation:

The quotient rule states that given a function `F(x) = (g(x))/(h(x)), (dF)/dx = ((dg)/(dx) * h(x) - g(x) * (dh)/dx)/(h(x))^2`

The power rule can help us find the derivatives of both numerator and denominator quickly.

`g(x) = 2x^3-1, (dg)/dx = 6x^2`
`h(x) = x^2, (dh)/dx = 2x, h^2(x)=x^4`

Thus...

`(dF)/dx = ((6x^2)(x^2) - (2x^3-1)(2x))/x^4 = (6x^4-4x^4+2x)/(x^4) = (2x^4+2x)/x^4 = 2 + 2/x^3`

Of note, the function will approach `-oo` as `x->0`

Answer 2

I would rewrite: `F(x) = 2x-1/x^2`

Explanation:

`f'(x) = 2+2/x^3 = (2x^3+2)/x^3`

using `d/dx(-1x^-2) = 2x^-3`

and rewriting as a single quotient.