2993 () posted 71 days ago by doubles 71 days ago by doubles +2994 / -1 66 comments share 66 comments share save hide report block hide child comments Comments (66) sorted by: top new old worst You're viewing a single comment thread. View all comments, or full comment thread. ▲ 18 ▼ – CykaSoaker76 18 points 70 days ago +18 / -0 f(x) = (16x^4 + 4x^2 + 1)(4x^3 - x) y1 = 16x^4 + 4x^2 + 1 y2 = 4x^3 - x y1' = 64x^3 + 8x y2' = 12x^2 - 1 f'(x) = (y1 * y2') + (y1' * y2) y1 * y2' = (16x^4 + 4x^2 + 1)(12x^2 - 1) = 192x^6 + 48x^4 + 12x^2 - 16x^4 - 4x^2 - 1 = 192x^6 + 32x^4 + 8x^2 - 1 y1' * y2 = (4x^3 - x)(64x^3 + 8x) = 256x^6 + 32x^4 - 64x^4 - 8x^2 = 256x^6 - 32x^4 - 8x^2 (y1 * y2') + (y1' * y2) = (192x^6 + 32x^4 + 8x^2 - 1) + (256x^6 - 32x^4 - 8x^2) = 448x^6 - 1 The math checks out! permalink save report block reply ▲ 2 ▼ – Jlin 2 points 70 days ago +4 / -2 I'm saddened this is not higher. permalink parent save report block reply
f(x) = (16x^4 + 4x^2 + 1)(4x^3 - x)
y1 = 16x^4 + 4x^2 + 1
y2 = 4x^3 - x
y1' = 64x^3 + 8x
y2' = 12x^2 - 1
f'(x) = (y1 * y2') + (y1' * y2)
y1 * y2' = (16x^4 + 4x^2 + 1)(12x^2 - 1)
= 192x^6 + 48x^4 + 12x^2 - 16x^4 - 4x^2 - 1
= 192x^6 + 32x^4 + 8x^2 - 1
y1' * y2 = (4x^3 - x)(64x^3 + 8x)
= 256x^6 + 32x^4 - 64x^4 - 8x^2
= 256x^6 - 32x^4 - 8x^2
(y1 * y2') + (y1' * y2) = (192x^6 + 32x^4 + 8x^2 - 1) + (256x^6 - 32x^4 - 8x^2)
= 448x^6 - 1
The math checks out!
I'm saddened this is not higher.