Find the equation of the tangent line....?

Question

(2x^3 - 1) / (x²) at the point (1,1)

i attempted it and my answer is too of from the answer in the book... so i cant go through and find a specific mistake >.< please help

Answer 1

it will be easier to find the derivative if you split it into two separate terms.

2x^3/x^2 - 1/x^2

2x - x^(-2)

The derivative is 2 + 2x^(-3)

If you substitute 1 in for x, the slope is 2 + 2(1)^-3 = 4

Using y = mx + b, replacing y and x with 1 and m with 4, you get
1 = 4*1 + b
So, b = -3

Tangent line is y = 4x - 3

Answer 2

quotient rule: low dee hi minus hi dee low over low squared:

[x²(6x²) - (2x³ - 1)2x]/x^4 =
(6x^4 - 4x^4 + 2x)/x^4 =
(2x^4 + 2x)/x^4 =
2 + 2/x³

slope of tangent at (1,1) is
2 + 2/(1)³ = 4
equation of tangent in point-slope form is
y-1 = 4(x-1), which is
y = 4x - 3 in slope-intercept form.

Answer 3

dy/dx=2+2/x^3
2+2/(1)^3=4
y=4x+c
1=4(1)+c
c= -3
y=4x-3