Jump to content
Main menu
Main menu
move to sidebar
hide
Navigation
Main Page
Community Portal
Babel user information
Root Category
Recent changes
States of WikiU
Help
Search
Search
English
Appearance
Donate
Create account
Log in
Personal tools
Donate
Create account
Log in
Pages for logged out editors
learn more
Contributions
Talk
Contents
move to sidebar
hide
Beginning
1
다항함수의 미분
Toggle 다항함수의 미분 subsection
1.1
증명
2
삼각함수의 미분
Toggle 삼각함수의 미분 subsection
2.1
증명
3
지수함수의 미분
Toggle 지수함수의 미분 subsection
3.1
증명
4
로그함수의 미분
Toggle 로그함수의 미분 subsection
4.1
증명
Toggle the table of contents
User
:
Hwangjy9/일변수함수의 미분 공식 증명
User page
Discussion
English
Read
Edit
View history
Tools
Tools
move to sidebar
hide
Actions
Read
Edit
View history
General
What links here
Related changes
User contributions
Logs
View user groups
Upload file
Special pages
Permanent link
Page information
Get shortened URL
Download QR code
Print/export
Create a book
Download as PDF
Printable version
Appearance
move to sidebar
hide
From Wikiversity
<
User:Hwangjy9
다음 공식들은 고등학교 수학 Ⅱ 과정에서 배우게 되는 미분 공식들을 모아놓은 것입니다.
다항함수의 미분
[
edit
]
(
x
n
)
′
=
n
x
n
−
1
{\displaystyle (x^{n})'=nx^{n-1}}
(단, n은 자연수)
증명
[
edit
]
(
x
n
)
′
=
lim
h
→
0
(
x
+
h
)
n
−
x
n
h
=
lim
h
→
0
(
x
+
h
−
x
)
(
(
x
+
h
)
n
−
1
+
x
(
x
+
h
)
n
−
2
+
⋯
+
x
n
−
2
(
x
+
h
)
+
x
n
−
1
)
h
=
n
x
n
−
1
{\displaystyle (x^{n})'=\lim _{h\to 0}{\frac {(x+h)^{n}-x^{n}}{h}}=\lim _{h\to 0}{\frac {(x+h-x)((x+h)^{n-1}+x(x+h)^{n-2}+\cdots +x^{n-2}(x+h)+x^{n-1})}{h}}=nx^{n-1}}
삼각함수의 미분
[
edit
]
(
sin
x
)
′
=
cos
x
{\displaystyle (\sin x)'=\cos x}
(
cos
x
)
′
=
−
sin
x
{\displaystyle (\cos x)'=-\sin x}
증명
[
edit
]
(
sin
x
)
′
=
lim
h
→
0
sin
(
x
+
h
)
−
sin
x
h
=
lim
h
→
0
2
cos
(
x
+
h
2
)
sin
h
2
h
=
cos
x
{\displaystyle (\sin x)'=\lim _{h\to 0}{\frac {\sin(x+h)-\sin x}{h}}=\lim _{h\to 0}{\frac {2\cos(x+{\frac {h}{2}})\sin {\frac {h}{2}}}{h}}=\cos x}
(
cos
x
)
′
=
lim
h
→
0
cos
(
x
+
h
)
−
cos
x
h
=
lim
h
→
0
−
2
sin
(
x
+
h
2
)
sin
h
2
h
=
−
sin
x
{\displaystyle (\cos x)'=\lim _{h\to 0}{\frac {\cos(x+h)-\cos x}{h}}=\lim _{h\to 0}{\frac {-2\sin(x+{\frac {h}{2}})\sin {\frac {h}{2}}}{h}}=-\sin x}
지수함수의 미분
[
edit
]
(
e
x
)
′
=
e
x
{\displaystyle (e^{x})'=e^{x}}
(
a
x
)
′
=
a
x
⋅
ln
a
{\displaystyle (a^{x})'=a^{x}\cdot \ln a}
증명
[
edit
]
(
e
x
)
′
=
lim
h
→
0
e
x
+
h
−
e
x
h
=
lim
h
→
0
e
x
(
e
h
−
1
)
h
=
e
x
{\displaystyle (e^{x})'=\lim _{h\to 0}{\frac {e^{x+h}-e^{x}}{h}}=\lim _{h\to 0}{\frac {e^{x}(e^{h}-1)}{h}}=e^{x}}
(
a
x
)
′
=
lim
h
→
0
a
x
+
h
−
a
x
h
=
lim
h
→
0
a
x
(
a
h
−
1
)
h
=
a
x
⋅
ln
a
{\displaystyle (a^{x})'=\lim _{h\to 0}{\frac {a^{x+h}-a^{x}}{h}}=\lim _{h\to 0}{\frac {a^{x}(a^{h}-1)}{h}}=a^{x}\cdot \ln a}
로그함수의 미분
[
edit
]
(
ln
x
)
′
=
1
x
{\displaystyle (\ln x)'={\frac {1}{x}}}
(
log
x
)
′
=
1
x
ln
a
{\displaystyle (\log x)'={\frac {1}{x\ln a}}}
증명
[
edit
]
(
ln
x
)
′
=
lim
h
→
0
ln
(
x
+
h
)
−
ln
x
h
=
lim
h
→
0
ln
(
1
+
h
x
)
h
x
⋅
x
=
1
x
{\displaystyle (\ln x)'=\lim _{h\to 0}{\frac {\ln(x+h)-\ln x}{h}}=\lim _{h\to 0}{\frac {\ln(1+{\frac {h}{x}})}{{\frac {h}{x}}\cdot x}}={\frac {1}{x}}}
(
log
x
)
′
=
lim
h
→
0
log
(
x
+
h
)
−
log
x
h
=
lim
h
→
0
log
(
1
+
h
x
)
h
x
⋅
x
=
1
x
ln
a
{\displaystyle (\log x)'=\lim _{h\to 0}{\frac {\log(x+h)-\log x}{h}}=\lim _{h\to 0}{\frac {\log(1+{\frac {h}{x}})}{{\frac {h}{x}}\cdot x}}={\frac {1}{x\ln a}}}