dy/dx = [x^1/2 + (x+1)^1/2]^2
thanks~!!
微積分 數學 HELP ~!!
2012-11-11 4:46 am
回答 (5)
2012-11-11 1:33 pm
✔ 最佳答案
dy/dx = [x^1/2 + (x+1)^1/2]^2=2[x^1/2 + (x+1)^1/2][1/2x^(-1/2) + 1/2(x+1)^(-1/2)]
=[x^1/2 + (x+1)^1/2][1/x^(1/2) + 1/(x+1)^(1/2)]
2012-11-13 3:27 am
dy/dx = [x^1/2 + (x+1)^1/2]^2
=2[x^1/2 + (x+1)^1/2][1/2x^(-1/2) + 1/2(x+1)^(-1/2)]
=[x^1/2 + (x+1)^1/2][1/x^(1/2) + 1/(x+1)^(1/2)]
=2[x^1/2 + (x+1)^1/2][1/2x^(-1/2) + 1/2(x+1)^(-1/2)]
=[x^1/2 + (x+1)^1/2][1/x^(1/2) + 1/(x+1)^(1/2)]
2012-11-11 8:13 pm
dy/dx = [x^1/2 + (x+1)^1/2]^2
By Chain Rule:
dy/dx = (dy/du)(du/dx)
Let [x^1/2 + (x+1)^1/2] be u
dy/du = (2)u^(2-1)
dy/du = 2u
du/dx = (1/2)x^[(1/2) -1]+(1/2)(x+1)^[(1/2) -1]
du/dx = (1/2)x^(-1/2)+(1/2)(x+1)^(-1/2)
∵ dy/dx = (dy/du)(du/dx)
∴dy/dx = [2u][(1/2)x^(-1/2)+(1/2)(x+1)^(-1/2)]
sub u=[x^1/2 + (x+1)^1/2] into dy/dx
dy/dx = 2[x^(1/2) + (x+1)^(1/2)][(1/2)x^(-1/2)+(1/2)(x+1)^(-1/2)]
dy/dx = 2[x^(1/2) + (x+1)^(1/2)][1/{2x^(1/2)} + 1/{2(x+1)^(1/2)} ]
2012-11-11 12:18:41 補充：
可以再簡化為:
dy/dx = [x^(1/2) + (x+1)^(1/2)][1/x^(1/2) + 1/(x+1)^(1/2)]
By Chain Rule:
dy/dx = (dy/du)(du/dx)
Let [x^1/2 + (x+1)^1/2] be u
dy/du = (2)u^(2-1)
dy/du = 2u
du/dx = (1/2)x^[(1/2) -1]+(1/2)(x+1)^[(1/2) -1]
du/dx = (1/2)x^(-1/2)+(1/2)(x+1)^(-1/2)
∵ dy/dx = (dy/du)(du/dx)
∴dy/dx = [2u][(1/2)x^(-1/2)+(1/2)(x+1)^(-1/2)]
sub u=[x^1/2 + (x+1)^1/2] into dy/dx
dy/dx = 2[x^(1/2) + (x+1)^(1/2)][(1/2)x^(-1/2)+(1/2)(x+1)^(-1/2)]
dy/dx = 2[x^(1/2) + (x+1)^(1/2)][1/{2x^(1/2)} + 1/{2(x+1)^(1/2)} ]
2012-11-11 12:18:41 補充：
可以再簡化為:
dy/dx = [x^(1/2) + (x+1)^(1/2)][1/x^(1/2) + 1/(x+1)^(1/2)]
參考: me
2012-11-11 7:57 am
題目呢= =看不到有題目
2012-11-11 6:47 am
你是要求y嗎????
收錄日期: 2021-04-24 10:08:10
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https://hk.answers.yahoo.com/question/index?qid=20121110000051KK00535