Forgot password
 Register account
View 1287|Reply 2

求偏导数

[Copy link]

6

Threads

7

Posts

0

Reputation

Show all posts

hjnk900 posted 2015-11-2 11:19 |Read mode
函数 f:R^3→R^3
Snip20151102_28.png

24

Threads

1017

Posts

46

Reputation

Show all posts

战巡 posted 2015-11-2 13:18
回复 1# hjnk900

这不基础题么?

易求
\[x=\frac{1}{2}\ln(\frac{u+v}{1+e^{2w}})+w\]
\[\frac{\partial x}{\partial u}=\frac{1}{2(u+v)}=\frac{1}{2(e^{2x}+e^{2y})}\]
\[\frac{\partial x}{\partial v}=\frac{1}{2(u+v)}=\frac{1}{2(e^{2x}+e^{2y})}\]
\[\frac{\partial x}{\partial w}=\frac{1}{1+e^{2w}}=\frac{1}{1+e^{2(x-y)}}\]

6

Threads

7

Posts

0

Reputation

Show all posts

original poster hjnk900 posted 2015-11-2 13:34
回复 2# 战巡

懂了 多谢!!!

Quick Reply

Advanced Mode
B Color Image Link Quote Code Smilies
You have to log in before you can reply Login | Register account

$\LaTeX$ formula tutorial

Mobile version

2025-7-21 11:47 GMT+8

Powered by Discuz!

Processed in 0.022692 seconds, 25 queries