calculus help please?

1 ⤊

calculus help please?

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

2007-12-31 15:05:48 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

2 answers

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 ]

2007-12-31 15:10:38 · answer #1 · answered by ¿ /\/ 馬 ? 7 · 1⤊ 1⤋

Hi,

f(x) = (3xÂ² -1)Â³ = (3xÂ²)Â³ - 3(3xÂ²)Â² + 3(3xÂ²) - 1
f(x) = 27x^6 - 27x^4 + 9xÂ² - 1

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

dÂ²y/dxÂ² = f''(x) = 810x^4 - 324xÂ² + 18

I hope that helps!! :-)

2007-12-31 16:13:42 · answer #2 · answered by Pi R Squared 7 · 0⤊ 1⤋