Derivative help please?

Question

Find the coefficient of the squared term in the simplified form for the second derivative, f "(x) for f(x) = (x^3 + 2x + 3)(3x^3 − 6x^2 − 8x + 1)

Answer 1

The function f(x) = (x³ + 2x + 3)(3x³ - 6x² - 8x + 1).

Simplify f(x).

f(x) = (x³ + 2x + 3)(3x³ - 6x² - 8x + 1)

= 3x⁶ + 6x⁴ + 9x³ - 6x⁵ - 12x³ - 18x²  - 8x⁴ - 16x² - 24x + x³ + 2x + 3

= 3x⁶ - 6x⁵ - 2x⁴ - 2x³ - 34x² - 22x + 3

⇒ f(x) = 3x⁶ - 6x⁵ - 2x⁴ - 2x³ - 34x² - 22x + 3.

Differentiate the function with resprct to x.

f '(x) = [3x⁶ - 6x⁵ - 2x⁴ - 2x³ - 34x² - 22x + 3]'

f '(x) = 3(x⁶)' - 6(x⁵)' - 2(x⁴)' - 2(x³)' - 34(x²)' - 22x' + 3'

Derivative of xⁿ = nx^{(n - 1)} and derivative of constant is zero.

f '(x) = 3(6x⁵) - 6(5x⁴) - 2(4x³) - 2(3x²) - 34(2x) - 22(1) + 0

f '(x) = 18x⁵ - 30x⁴ - 8x³ - 6x² - 68x - 22.

Again differentiate the above equation with respect to x.

f ''(x) = [18x⁵ - 30x⁴ - 8x³ - 6x² - 68x - 22]'

f ''(x) = 18(x⁵)' - 30(x⁴)' - 8(x³)' - 6(x²)' - 68x' - 22'

f ''(x) = 18(5x⁵) - 30(4x³) - 8(3x²) - 6(2x) - 68(1) - 0

f ''(x) = 90x⁵ - 120x³ - 24x² - 12x - 68.

Therefore, coefficient of the squared term(i.e, coefficient of x²) of  f ''(x) is - 24.