# Thread: second derivative

1. ## second derivative

f(x) = (3x+1)^3

f'(x) = 3(3x+1)^2(3)
f'(x) = 9(3x+1)^2

f"(x) = 9[(2)(3x+1)(3)] + [9(3x+1)^2]
f"(x) = 54(3x+1) + 9(3x+1)^2
f"(x) = 9(3x+1)[6 + (3x+1)^2]
f"(x) = 9(3x+1)(7+3x)

Am I correct?

2. Originally Posted by startingover
f(x) = (3x+1)^3

f'(x) = 3(3x+1)^2(3)
f'(x) = 9(3x+1)^2

f"(x) = 9[(2)(3x+1)(3)] + [9(3x+1)^2]
f"(x) = 54(3x+1) + 9(3x+1)^2
f"(x) = 9(3x+1)[6 + (3x+1)^2]
f"(x) = 9(3x+1)(7+3x)

Am I correct?
What's going on in line 4? All you need to do is:
f''(x) = 9 * 2(3x + 1)*3 = 54(3x + 1) = 162x + 54

-Dan

3. Originally Posted by startingover
f(x) = (3x+1)^3

f'(x) = 3(3x+1)^2(3)
f'(x) = 9(3x+1)^2

f"(x) = 9[(2)(3x+1)(3)] + [9(3x+1)^2]
f"(x) = 54(3x+1) + 9(3x+1)^2
f"(x) = 9(3x+1)[6 + (3x+1)^2]
f"(x) = 9(3x+1)(7+3x)

Am I correct?
You are right until you decided to take the second derivative:

f(x) = (3x+1)^2
f'(x) = 3(3x+1)^2*(3)

Now in order to get f''(x) you must use the product rule (not the triple product rule as constants hold over) thus:

f''(x) = (3)*(2(3x+1))*(3)*(3)

You do not have to but I choose to simplify so

f''(x)= 54(3x+1)

4. Originally Posted by qbkr21
You are right until you decided to take the second derivative:

f(x) = (3x+1)^2
f'(x) = 3(3x+1)^2*(3)

Now in order to get f''(x) you must use the product rule (not the triple product rule as constants hold over) thus:

f''(x) = (3)*(2(3x+1))*(3)*(3)

You do not have to but I choose to simplify so

f''(x)= 54(3x+1)
Product rule? Why not just take the derivative of (3x + 1)^2 using a combination of the chain and power rules? (Like startingover did when (s)he took the derivative of (3x + 1)^3 in the second line.) Thinking of this in terms of the product rule is overkill.

-Dan

5. ## Re:

Good point! Sorry for the overkill...