Differentiate each of the following with respect to x.
13. y = (2x^2 +1) [ (2x^2+1)^1/2 ]
15. y = 2x [(2x+1)^3]^1/2
M2 DIFFERNTIATION
2010-10-31 7:29 am
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2010-10-31 9:14 am
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參考: My Math knowledge
2010-10-31 8:26 pm
13. y = (2x^2 +1) [ (2x^2+1)^1/2 ]
dy/dx=(2x^2+1)[d/dx(2x^2+1)^1/2)]+[(2x^2+1)^1/2][d/dx(2x^2+1)]
=(2x^2+1)(1/2)[(2x^+1)^(-1/2)](4x)+[(2x^2+1)^1/2](4x)
=[(2x^2+1)^(1/2)](1/2)(4x)+[(2x^2+1)^1/2](4x)
=[(2x^2+1)^(1/2)](4x)[(1/2)+1]
=[(2x^2+1)^(1/2)](4x)(3/2)
=6x[(2x^2+1)^(1/2)]
2010-10-31 12:29:37 補充：
Q15 answer also can express to 2(5x+1)[(2x+1)^(1/2)]
dy/dx=(2x^2+1)[d/dx(2x^2+1)^1/2)]+[(2x^2+1)^1/2][d/dx(2x^2+1)]
=(2x^2+1)(1/2)[(2x^+1)^(-1/2)](4x)+[(2x^2+1)^1/2](4x)
=[(2x^2+1)^(1/2)](1/2)(4x)+[(2x^2+1)^1/2](4x)
=[(2x^2+1)^(1/2)](4x)[(1/2)+1]
=[(2x^2+1)^(1/2)](4x)(3/2)
=6x[(2x^2+1)^(1/2)]
2010-10-31 12:29:37 補充：
Q15 answer also can express to 2(5x+1)[(2x+1)^(1/2)]
收錄日期: 2021-04-13 17:36:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101030000051KK01720