Help just Calculus 1 here?

Question

using definition to take derivative of 3x^2 - 2x (answer of course is 6x - 2) but if I take a limit of a secant line I got - (6x + 2)
Someone please help and show me step by step I checked my algebra still cant see where I got it wrong!

Answer 1

The definition of the derivative is lim(d->0) [f(x+d) - f(x)]/d. In your problem, f(x+d) = 3*(x+d)^2 - 2*(x+d); then f(x+d)-f(x) is

3(x+d)^2 - 2*(x+d) - 3x^2 + 2x. Expand the terms to get

3x^2 +6xd +3d^2 - 2x -2d -3x^2 +2x; simplify, cancelling terms

6xd +3d^2 -2d. Now divide by d as in the definition to get

6x +3d -2. The limit of this as d->0 is 6x - 2

Answer 2

secant = Δy/Δx
=[f(c+Δx) - f(c)] / [c+Δx - c]
=[f(c+Δx) - f(c)] / Δx

={[3(c+Δx)^2 - 2(c+Δx)] - [3c^2 - 2c]}/Δx

= [ 3(c^2 +2cΔx + Δx^2) - 2c - 2Δx - 3c^2 + 2c]/Δx

= [ 6cΔx + 6Δx^2 - 2Δx] / Δx

= 6c + 6Δx - 2
lim = 6c - 2
Δx --> 0