Expert: Sherman D. - 10/7/2005

Question
find the first and second derivatives.

1) y=x^2+x+8
2) y=(4x^3/3) - 4

Answer
1.)
y = x^2 + x + 8
y' = (x^2)' + (x)' + 8
Derivatives of Constants is 0
y' = (x^2)' + (x)'
The power rule states that x^n is nx^(n - 1)
y' = (x^(2 - 1)) + (x^(1 - 1))
y' = (x^1) + x^0
y' = 2x + 1

so y' = 2x + 1

y' = 2x + 1
y" = (2x^(1 - 1)) + 0
y" = 2x^0
y" = 2 * 1
y" = 2

y' = 2x + 1
y" = 2

2.)
y = (4/3)x^3 - 4
y' = (4/3)x^3
y' = ((4/3) * 3)x^(3 - 1)
y' = 4x^2

y' = 4x^2
y" = 4(x^2)
y" = (4 * 2)(x^(2 - 1))
y" = 8x^1
y" = 8x

y' = 4x^2
y" = 8x

Since i haven't done derivatives in awhile, these answers may be subject to change. I asked someone else and i am waiting on a reply. If you don't here back from me within the week, they are correct.