Find the second derivative of the function.

f(x)= (3x^3+7)^7

no problem

f' = 7(3x^3+7)^6 * 9x^2
= 63x^2 (3x^3+7)^6

f" = 126x(3x^3+7)^6 + 63x^2 * 6(3x^3+7)^5 * 9x^2
= 126x(3x^3+7)^5 (3x^3+7 + 27x^3)
= 126x(3x^3+7)^5 (30x^3+7)

I have these ans choices:

A. f= 63x(7+3x^3)^5 (14+60x^3)

B. f= 63x(7+3x^3)^5 (14+63x^3)

C. f= 63x(7+3x)^5 (14+63x^3)

D. f= 63x(7+3x^3)^5 (14-60x^3)

Come on, guy. move the 2 from 126 into the last term. Algebra I, here.

good man.

no idea why they didn't factor out the 2 as well.

To find the second derivative of the function f(x) = (3x^3 + 7)^7, we will need to use the chain rule twice. Here's how you can do it step by step:

Step 1: Find the first derivative.
Let's start by finding the first derivative of f(x) with respect to x.
f'(x) = 7(3x^3 + 7)^6 * d/dx(3x^3 + 7)

Step 2: Differentiate the inner function.
To find d/dx(3x^3 + 7), we differentiate the expression inside the parentheses as follows:
d/dx(3x^3 + 7) = 9x^2

Step 3: Substitute the derivative into the first derivative.
Now, substitute d/dx(3x^3 + 7) = 9x^2 into the first derivative expression:
f'(x) = 7(3x^3 + 7)^6 * 9x^2

Step 4: Simplify the expression.
Multiply the coefficients and combine the exponents:
f'(x) = 63x^2(3x^3 + 7)^6

Step 5: Find the second derivative.
To find the second derivative, we need to differentiate f'(x) with respect to x:
f''(x) = d/dx [63x^2(3x^3 + 7)^6]

Step 6: Use the chain rule again.
Apply the chain rule to the expression 63x^2(3x^3 + 7)^6:
f''(x) = 63x^2 * d/dx[(3x^3 + 7)^6]

Step 7: Differentiate the inner function.
Differentiate (3x^3 + 7)^6 with respect to x using the chain rule:
d/dx[(3x^3 + 7)^6] = 6(3x^3 + 7)^5 * d/dx(3x^3 + 7)

Step 8: Differentiate the inner function again.
Differentiate 3x^3 + 7 with respect to x:
d/dx(3x^3 + 7) = 9x^2

Step 9: Substitute the derivative into the second derivative.
Substitute d/dx(3x^3 + 7) = 9x^2 and simplify the expression:
f''(x) = 63x^2 * 6(3x^3 + 7)^5 * 9x^2

Step 10: Simplify the expression.
Multiply the coefficients and combine the exponents:
f''(x) = 3402x^4(3x^3 + 7)^5

Therefore, the second derivative of f(x) = (3x^3 + 7)^7 is f''(x) = 3402x^4(3x^3 + 7)^5.