都要用 隱含數微分 2√x-√y=1 二階微分 (√x-1)/(y+1)=y 一階微分 要有過程~感恩^^?
回答 (2)
2016-01-15 11:25 am
2√x-√y=1
2√x-1=√y
y=(2√x-1)^2
y'=2(2√x-1)(x^-0.5)
y"=2(-0.5x^-1.5)(x^-0.5)
y"=-x^(-2)
(√x-1)/(y+1)=y
√x-1=y(y+1)
0.5x^(-0.5)=(dy/dx)(y+1)+(dy/dx)(y)
(2y)(dy/dx)+dy/dx=0.5x^(-0.5)
dy/dx=(0.5x^-0.5)/(2y+1)
2√x-1=√y
y=(2√x-1)^2
y'=2(2√x-1)(x^-0.5)
y"=2(-0.5x^-1.5)(x^-0.5)
y"=-x^(-2)
(√x-1)/(y+1)=y
√x-1=y(y+1)
0.5x^(-0.5)=(dy/dx)(y+1)+(dy/dx)(y)
(2y)(dy/dx)+dy/dx=0.5x^(-0.5)
dy/dx=(0.5x^-0.5)/(2y+1)
2016-01-14 12:12 am
(1) 隱含數微分 2√x-√y=1 二階微分
1/√x - y'/2√y = 0
y' = 2√y/√x
y" = 2{√x * y'/2√y - √y * 1/√x}/x
= {√x/√y * (2√y/√x) - 2√y/√x}/x
= 2*(1 - √y/√x)/x
= 2 * (√x - √y) / x√x
(2) (√x-1)/(y+1)=y 一階微分 要有過程~感恩^?
√x - 1 = y(y+1) = y^2 + y
1/2√x = 2yy' + y' = (2y + 1)y'
y' = 1 / 2(2y + 1)√x
1/√x - y'/2√y = 0
y' = 2√y/√x
y" = 2{√x * y'/2√y - √y * 1/√x}/x
= {√x/√y * (2√y/√x) - 2√y/√x}/x
= 2*(1 - √y/√x)/x
= 2 * (√x - √y) / x√x
(2) (√x-1)/(y+1)=y 一階微分 要有過程~感恩^?
√x - 1 = y(y+1) = y^2 + y
1/2√x = 2yy' + y' = (2y + 1)y'
y' = 1 / 2(2y + 1)√x
收錄日期: 2021-04-18 14:16:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160109001653AAMxIgq