dkfr.net
当前位置:首页 >> xy E x 2y 的导数 >>

xy E x 2y 的导数

(e^xy)' =[(e^x)^y]' =[(e^x)^y]ln(e^x) =[(e^xy]*x =xe^xy

xy=e^(x+y) 两边求导: y + xy ′ = e^(x+y) * (1+y ′) y + xy ′ = e^(x+y) + e^(x+y) * y ′ xy ′ - e^(x+y) * y ′ = e^(x+y) - y y ′ = {e^(x+y) - y} / { x - e^(x+y) } ================================= xy=e^x+y 两边求导: y + xy ′ = e^x...

隐函数求导如下: 方程两边求导: y+xy'=e^(x+y)(1+y') y+xy'=e^(x+y)+y'e^(x+y) y'[x-e^(x+y)]=e^(x+y)-y y'=[e^(x+y)-y]/[x-e^(x+y)].

若:e^(xy) = c ----- (0) 问题为隐函数求导 两边对x求导: e^(xy) (y+xy') = 0 y+xy' = 0 y' = -y/x ---------------------- (1) xy = ln c ------------------------(2) y = lnc / x -----------------------(3) y' = - lnc / x² -------...

xy=e^(x+y) 两边对x求导得 y+xy'=e^(x+y)(1+y') [x-e^(x+y)]y'=e^(x+y)-y y'=[e^(x+y)-y]/[x-e^(x+y)]

方程xy=e^(x+y)确定的隐函数y的导数: y'=[e^(x+y)-y]/[x-e^(x+y)] 解题过程: 方程两边求导: y+xy'=e^(x+y)(1+y') y+xy'=e^(x+y)+y'e^(x+y) y'[x-e^(x+y)]=e^(x+y)-y 得出最终结果为: y'=[e^(x+y)-y]/[x-e^(x+y)] 隐函数求导方法: 1.先把隐...

两边同时对x求导 e^x-e^yy'=cos(xy)(y+xy') y'=[e^x-ycos(xy)]/【xcos(xy)+e^y】 dy/dx=[e^x-ycos(xy)]/【xcos(xy)+e^y】

网站首页 | 网站地图
All rights reserved Powered by www.dkfr.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com