The derivative of (x^3-3x^2+4)/x^2?

Question

I just started calculus and i know this is a pretty simple question just mostly using the power rule but i can't seem to get the right answer which is (x^3-8)/x^3

I was hoping if someone could show me step by step what they did because i have done this problem a few times and cannot figure it out

it seems like people are factoring out the x^2 from the numerator and canceling but how can you do that the +4 does not have a common factor

Answer 1

y=(x³ - 3x² + 4)/x²

y=x - 3 + 4/x²

dy/dx= 1 - 0 - 8/x³

dy/dx= 1 - 8/x³

n_n

Answer 2

(x^3 - 3x^2 + 4)/x^2 =

take out x^2 common from numerator so that you cancel it with denominator

x^2(x - 3 + 4/x^2)/x^2 =

x - 3 + 4 x^(-2)

difeerentiate with respect to x

1 + 4(-2)x^(-3) =

1 - 8/x^3 =

(x^3 - 8)/x^3

Answer 3

One way to solve this one is to distribute your 1/x^2 term over each term in the numerator before differentiating. This will give you

x^3/x^2 - 3x^2/x^2 + 4/x^2, which simplifies to

x - 3 + 4x^(-2), easy enough to differentiate to

1 + 4(-2)x^(-3) (the constant term -3 disappears during the differentiation)
= 1 - 8x^(-3), which may be rewritten as
(x^3 - 8) / x^3

Answer 4

y = (x^3-3x^2+4)/x^2
= x - 3 + 4/x^2

y' = 1 - 8/x^3 = (x^3-8)/x^3

Answer 5

[x^2(3x^2 - 6x) - (x^3 - 3x^2 +4)(2x)] /(x^2)^2
[3x^4 - 6x^3 - 2x^4 +6x^3 - 8x] / x^4
(x^4 - 8x)/x^4
x(x^3 - 8) / x4 = (x^3 - 8)/x^3
just remember theorem 3.8 the Quotient Rule- bottom x derivative of the top - the top x the derivative of bottom.

Answer 6

f' = 1/x^2*(3x^2 -6x) -2(x^3-3x^2+4)/x^3

f' =3 -6/x -2 + 6/x -8/x^3 =1-8/x^3 = (x^3 - 8)/x^3

Answer 7

Good luck.