求d x,ln x怎么求

首页 > 大全 > 作者:YD1662022-12-07 00:08:38

dⅹ²/dⅹ=2ⅹ

dy²/dⅹ,令μ=y²→

dμ/dx=(dμ/dy)(dy/dⅹ)=2y(dy/dⅹ)

ⅹ² y²=4→

dⅹ²/dⅹ dy²/dⅹ=d4/dⅹ→

2ⅹ 2ydy/dⅹ=0→

dy/dⅹ=-ⅹ/y

点A(1,√3)为该圆上的点,则该点斜率为dy/dⅹ=-ⅹ/y=-1/√3→

该点切线为(y-√3)/(ⅹ-1)=-1/√3,→y=-(√3/3)x 4√3/3

2y siny=(x²/π) 1→

d(2y)/dⅹ d(siny)/dⅹ=

d(ⅹ²/π)/dⅹ d(1)/dⅹ→

2dy/dⅹ cosydy/dⅹ=2ⅹ/π→

d(2dy/dⅹ)/dⅹ

d(cosydy/dⅹ)/dⅹ=d(2ⅹ/π)/dⅹ

→2d²y/dⅹ² ds/dⅹ=2/π①

s=cosydy/dⅹ

令u=cosy,m=dy/dⅹ→s=um

ds/dⅹ=mdu/dⅹ udm/dⅹ=

(dy/dⅹ)(du/dⅹ) cosyd²y/dⅹ²

du/dⅹ=dcosy/dⅹ=

(dcosy/dy)(dy/dx)=-sinydy/dⅹ→

ds/dⅹ=

-siny(dy/dⅹ)² cosyd²y/dⅹ²→①为

d²y/dⅹ²(2 cosy) -siny(dy/dⅹ)²=

π/2(d²y/dⅹ²与(dy/dⅹ)²不同)

栏目热文

文档排行

本站推荐

Copyright © 2018 - 2021 www.yd166.com., All Rights Reserved.