求下列各函數的dy/dx，拜託高手解答
1. y=(ln x)^2 (用^2表示平方)
2. 由函數關係 y+ln xy=1，利用隱函數微分求出 dy/dx
微積分 dy/dx求解?
2016-06-16 5:51 am
回答 (2)
2016-06-16 8:34 am
✔ 最佳答案
1.dy / dx
= 2 * ln x * ( ln x )'
= 2 * ln x * (1/x)
= 2 ( ln x ) / x ..... Ans
2.
令 F(x,y) = y + ( ln xy ) - 1 = y -1 + ln x + ln y
Fx = ∂F/∂x = 1/x
Fy = ∂F/∂y = 1 + 1/y = (y+1)/y
dy / dx
= - Fx / Fy
= - 1/x / [ (y+1)/y ]
= - y / [ x(y+1) ] ..... Ans
2016-06-27 2:38 pm
1.
y=(ln x)^2
dy/dx=2(In x)(1/x)
dydx=(2Inx)/x
2.
y+(In x)+(In y)=1
dy/dx +(1/x)+(1/y)(dy/dx)=0
dy/dx(1-1/y)=-1/x
dy/dx=-y/[x(y-1)]
y=(ln x)^2
dy/dx=2(In x)(1/x)
dydx=(2Inx)/x
2.
y+(In x)+(In y)=1
dy/dx +(1/x)+(1/y)(dy/dx)=0
dy/dx(1-1/y)=-1/x
dy/dx=-y/[x(y-1)]
收錄日期: 2021-04-18 15:08:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160615215125AA64rlS