dy/dx (x^2-2x+6)^3 (x^2+4x-5)^4
回答 (2)
2008-05-30 9:45 pm
呢一ｄ要自己答的嗎！！！
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佢地唔會教你的
你一係就問下老師
你在這ｄ人都只係比個ａｎｓ你
佢地唔會教你的
參考: Ｉ ＴＨＩＮＫ Ｉ ＭＡＹ ＨＥＬＰ ＹＯＵ ＾＾
2008-05-27 5:21 pm
Let y= (x^2-2x+6)^3 (x^2+4x-5)^4
By Product and Chain Rule:
dy/dx = 3(x^2-2x+6)^2 * (2x-2) * (x^2+4x-5)^4 + (x^2-2x+6)^3 * 4(x^2+4x-5)^3 * (2x +4)
=(x^2-2x+6)^2(x^2+4x-5)^3 * [6(x-1)(x^2+4x-5) + 8(x^2-2x+6)(x+2) ]
=(x^2-2x+6)^2(x^2+4x-5)^3 * 2[7x^3 + 8x^2 -19x + 63]
By Product and Chain Rule:
dy/dx = 3(x^2-2x+6)^2 * (2x-2) * (x^2+4x-5)^4 + (x^2-2x+6)^3 * 4(x^2+4x-5)^3 * (2x +4)
=(x^2-2x+6)^2(x^2+4x-5)^3 * [6(x-1)(x^2+4x-5) + 8(x^2-2x+6)(x+2) ]
=(x^2-2x+6)^2(x^2+4x-5)^3 * 2[7x^3 + 8x^2 -19x + 63]
收錄日期: 2021-04-15 15:19:13
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080527000051KK00379