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^^?

2016-01-09 8:16 am

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)
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


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