find y' where y is gien by:
1) cos2t / sin3t
2) t+sint / 1+ e^t
3) √x / x+1
some A.maths chain rule quester
2008-07-15 4:10 am
回答 (1)
2008-07-15 4:47 am
✔ 最佳答案
1) y' = [(-2sin2t)sin3t-cos2t(3cos3t)]/ sin^2 (3t)= -( 2sin2tsin3t+3cos2tcos3t)/sin^2 (3t)
2) y' = [(1+cost)(1+e^t)-e^t(t+sint)]/(1+e^t)^2
= [1 +cost+ e^t(1+cost -t-sint)] / (1+e^t)^2
3) y' = [1/(2√x) *(x+1)- √x] /(x+1)^2
= (1-x)/2√x(x+1)^2
收錄日期: 2021-04-13 15:49:05
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