Find the derivative of 2x^3 - x?

Question

using Lim
h->0

Answer 1

y = 2x^3 - x

y' = lim h->0 [f(x+h)-f(x)]/h
= lim h->0 {[2(x+h)^3 - (x+h)] - [2x^3 - x]} /h
= lim h->0 {[2(x^3 + 3x^2h+3xh^2 + h^3) - x - h] - 2x^3 + x}/h
= lim h->0 { 2x^3 + 6x^2h+6xh^2 + 2h^3 - x - h - 2x^3 + x}/h
= lim h->0 { 6x^2h+6xh^2 + 2h^3 - h} / h
= lim h->0 { h( 6x^2+6xh + 2h^2 - 1)} / h
= lim h->0 ( 6x^2+6xh + 2h^2 - 1)
= 6x^2 - 1

Answer 2

Procedure to get derivative of:
ax^b:
a(b) x^(b-1)

so, derivatives of:
2x^3 = 2(3) x^(3-1) = 6x^2
and
x = 1(1) x^(1-1) = x^0 = 1

which comes out to derivative of::
2x^3 - x = 6x^2 - 1

Ciao!

Answer 3

substitute (x+h) for x in the expression and multiply out. You will have 2(x+h)^3-(x+h).
Now subtract out the initial function. So you will have 2(x+h)^3-2x^3-(x+h)+(x). In the cubic term, the x^3 will drop out, leaving behind terms in x^2h, xh^2 and h^3, while the -x term will yield -h.
Now we divide by h, leaving terms with x^2, xh and h^2 and -1. Now you let h approach 0. Any term with h in it will drop out.

Answer 4

y = 2x ³ - x
y ' = 6 x ² - 1

Answer 5

6x^2 - 1

Answer 6

y=2x^3 - x
Δy+y=2(x+Δx)^3-(x+Δx)
Δy=2(x+Δx)^3-(x+Δx)-(2x^3 - x)
Δy=2(x^3+3x^2Δx+3xΔx^2+Δx^3)-x-Δx-2x^3+x
Δy=2x^3+6x^2Δx+6xΔx^2+2Δx^3-x-Δx-2x^3+x
Δy=6x^2Δx+xΔx^2+2Δx^3-Δx
Δy=Δx(6x^2+Δx+2Δx^2-1)
Δy/Δx=6x^2+Δx+2Δx^2-1
limit of Δy/Δx as Δx approaces 0
Δy/Δx=6x^2-1