Question

How do you find the derivative of `[(2x^3) - (4x^2) + 3] / x^2`?

Answer 1

`d/dx({2x^3-4x^2+3}/x^2)={2(x^3-3)}/x^3`

Explanation:

We use the quotient rule, which says for `f(x)=g(x)/{h(x)}, f'(x)={h(x)g'(x)-h'(x)g(x)}/{(h(x))^2}`.

So, we first let `f(x)={2x^3-4x^2+3}/x^2`.

Thus, `g(x)=2x^3-4x^2+3` and `h(x)=x^2`.

Next, we find `g'(x)` and `h'(x)`

This just involves the power rule, so
`g'(x)=3*2x^2-2*4x+0=6x^2-8x`
`h'(x)=2x`

Now, we just plug these values into the quotient rule to get the following:
`f'(x)={(x^2)(6x^2-8x)-(2x)(2x^3-4x^2+3)}/{(x^2)^2}`
`f'(x)={(6x^4-8x^3)-(4x^4-8x^3+6x)}/{x^4}`
`f'(x)={6x^4-4x^4-8x^3+8x^3-6x}/{x^4}`
`f'(x)={2x^4-6x}/{x^4}`
`f'(x)={2x(x^3-3)}/{x^4}`
`f'(x)={2(x^3-3)}/x^3`
Therefore, `d/dx({2x^3-4x^2+3}/x^2)={2(x^3-3)}/x^3`