跳转到内容
Main menu
Main menu
move to sidebar
hide
Navigation
Main Page
Help
Browse
Cookbook
Wikijunior
Featured books
Recent changes
Random book
Using Wikibooks
Community
Reading room forum
Community portal
Bulletin Board
Help out!
Policies and guidelines
Contact us
Search
Search
Donations
Appearance
Create account
Log in
Personal tools
Create account
Log in
Pages for logged out editors
learn more
Contributions
Discussion for this IP address
微积分/链式法则/解答
Add languages
Add links
Book
Discussion
English
Read
Edit
Edit source
View history
Tools
Tools
move to sidebar
hide
Actions
Read
Edit
Edit source
View history
General
What links here
Related changes
Upload file
Special pages
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Sister projects
Wikipedia
Wikiversity
Wiktionary
Wikiquote
Wikisource
Wikinews
Wikivoyage
Commons
Wikidata
MediaWiki
Meta-Wiki
Print/export
Create a collection
Download as PDF
Printable version
In other projects
外观
移至侧边栏
隐藏
来自维基教科书,开放的书籍,构建开放的世界
<
微积分
|
链式法则
1. 计算
f
′
(
x
)
{\displaystyle f'(x)}
,如果
f
(
x
)
=
(
x
2
+
5
)
2
{\displaystyle f(x)=(x^{2}+5)^{2}}
,首先展开并直接求导,然后对
f
(
u
(
x
)
)
=
u
2
{\displaystyle f(u(x))=u^{2}}
应用链式法则,其中
u
=
x
2
+
5
{\displaystyle u=x^{2}+5}
。比较结果。
第一种方法
f
(
x
)
=
x
4
+
10
x
2
+
25
{\displaystyle f(x)=x^{4}+10x^{2}+25}
f
′
(
x
)
=
4
x
3
+
20
x
{\displaystyle \mathbf {f'(x)=4x^{3}+20x} }
第二种方法
f
′
(
u
(
x
)
)
=
d
f
d
u
⋅
d
u
d
x
=
2
u
⋅
2
x
=
2
(
x
2
+
5
)
⋅
2
x
=
4
x
3
+
20
x
{\displaystyle f'(u(x))={\frac {df}{du}}\cdot {\frac {du}{dx}}=2u\cdot 2x=2(x^{2}+5)\cdot 2x=\mathbf {4x^{3}+20x} }
两种方法得到相同的结果。
第一种方法
f
(
x
)
=
x
4
+
10
x
2
+
25
{\displaystyle f(x)=x^{4}+10x^{2}+25}
f
′
(
x
)
=
4
x
3
+
20
x
{\displaystyle \mathbf {f'(x)=4x^{3}+20x} }
第二种方法
f
′
(
u
(
x
)
)
=
d
f
d
u
⋅
d
u
d
x
=
2
u
⋅
2
x
=
2
(
x
2
+
5
)
⋅
2
x
=
4
x
3
+
20
x
{\displaystyle f'(u(x))={\frac {df}{du}}\cdot {\frac {du}{dx}}=2u\cdot 2x=2(x^{2}+5)\cdot 2x=\mathbf {4x^{3}+20x} }
两种方法得到相同的结果。
2. 利用链式法则,令
y
=
u
{\displaystyle y={\sqrt {u}}}
和
u
=
1
+
x
2
{\displaystyle u=1+x^{2}}
,计算
y
=
1
+
x
2
{\displaystyle y={\sqrt {1+x^{2}}}}
的导数。
:
d
y
d
u
=
1
2
u
;
d
u
d
x
=
2
x
{\displaystyle {\frac {dy}{du}}={\frac {1}{2{\sqrt {u}}}};\quad {\frac {du}{dx}}=2x}
d
y
d
x
=
d
y
d
u
⋅
d
u
d
x
=
1
2
1
+
x
2
⋅
2
x
=
x
1
+
x
2
{\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}\cdot 2x=\mathbf {\frac {x}{\sqrt {1+x^{2}}}} }
:
d
y
d
u
=
1
2
u
;
d
u
d
x
=
2
x
{\displaystyle {\frac {dy}{du}}={\frac {1}{2{\sqrt {u}}}};\quad {\frac {du}{dx}}=2x}
d
y
d
x
=
d
y
d
u
⋅
d
u
d
x
=
1
2
1
+
x
2
⋅
2
x
=
x
1
+
x
2
{\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}\cdot 2x=\mathbf {\frac {x}{\sqrt {1+x^{2}}}} }
导航
:
主页
·
预备微积分
·
极限
·
微分
·
积分
·
参数方程和极坐标方程
·
数列与级数
·
多元微积分
·
扩展
·
参考文献
分类
:
书籍:微积分
华夏公益教科书