calculus help please?

Question

Find d^2y/dx^2 for y=(3x^2-1)^3

Answer 1

y=(3x^2-1)^3

Use the chain rule for (3x^2-1)^3

y'=3(3x^2-1)^2 *6x
y'=18x(3x^2-1)^2

y''= 18(3x^2-1)^2 + 18x(2(3x^2-1) *6x)
y''= 18(3x^2-1)^2 + 216x^2(3x^2-1)
y''= 18(3x^2-1) [ (3x^2-1) + 12x^2 ]

Answer 2

Hi,

f(x) = (3x² -1)³ = (3x²)³ - 3(3x²)² + 3(3x²) - 1
f(x) = 27x^6 - 27x^4 + 9x² - 1

dy/dx = f'(x) = 162x^5 - 108x^3 + 18x

d²y/dx² = f''(x) = 810x^4 - 324x² + 18

I hope that helps!! :-)