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微積分/微分/微分的定義/解答
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<
微積分
|
微分
|
微分的定義
1. 求曲線
y
=
x
2
{\displaystyle y=x^{2}}
在點
(
1
,
1
)
{\displaystyle (1,1)}
處的切線的斜率。
函式
f
{\displaystyle f}
在
x
0
{\displaystyle x_{0}}
處的斜率的定義是
lim
h
→
0
[
f
(
x
0
+
h
)
−
f
(
x
0
)
h
]
{\displaystyle \lim _{h\to 0}\left[{\frac {f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}}\right]}
代入
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
和
x
0
=
1
{\displaystyle x_{0}=1}
得到
lim
h
→
0
[
(
1
+
h
)
2
−
1
h
]
=
lim
h
→
0
[
h
2
+
2
h
h
]
=
lim
h
→
0
[
h
(
h
+
2
)
h
]
=
lim
h
→
0
h
+
2
=
2
{\displaystyle {\begin{aligned}\lim _{h\to 0}\left[{\frac {(1+h)^{2}-1}{h}}\right]&=\lim _{h\to 0}\left[{\frac {h^{2}+2h}{h}}\right]\\&=\lim _{h\to 0}\left[{\frac {h(h+2)}{h}}\right]\\&=\lim _{h\to 0}h+2\\&=\mathbf {2} \end{aligned}}}
函式
f
{\displaystyle f}
在
x
0
{\displaystyle x_{0}}
處的斜率的定義是
lim
h
→
0
[
f
(
x
0
+
h
)
−
f
(
x
0
)
h
]
{\displaystyle \lim _{h\to 0}\left[{\frac {f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}}\right]}
代入
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
和
x
0
=
1
{\displaystyle x_{0}=1}
得到
lim
h
→
0
[
(
1
+
h
)
2
−
1
h
]
=
lim
h
→
0
[
h
2
+
2
h
h
]
=
lim
h
→
0
[
h
(
h
+
2
)
h
]
=
lim
h
→
0
h
+
2
=
2
{\displaystyle {\begin{aligned}\lim _{h\to 0}\left[{\frac {(1+h)^{2}-1}{h}}\right]&=\lim _{h\to 0}\left[{\frac {h^{2}+2h}{h}}\right]\\&=\lim _{h\to 0}\left[{\frac {h(h+2)}{h}}\right]\\&=\lim _{h\to 0}h+2\\&=\mathbf {2} \end{aligned}}}
2. 利用導數的定義求函式
f
(
x
)
=
2
x
+
3
{\displaystyle f(x)=2x+3}
的導數。
f
′
(
x
)
=
lim
Δ
x
→
0
(
2
(
x
+
Δ
x
)
+
3
)
−
(
2
x
+
3
)
Δ
x
=
lim
Δ
x
→
0
2
x
+
2
Δ
x
+
3
−
2
x
−
3
)
Δ
x
=
lim
Δ
x
→
0
2
Δ
x
Δ
x
=
lim
Δ
x
→
0
2
=
2
{\displaystyle {\begin{aligned}f^{'}(x)&=\lim _{\Delta x\to 0}{\frac {(2(x+\Delta x)+3)-(2x+3)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x+2\Delta x+3-2x-3)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0}2\\&=\mathbf {2} \end{aligned}}}
f
′
(
x
)
=
lim
Δ
x
→
0
(
2
(
x
+
Δ
x
)
+
3
)
−
(
2
x
+
3
)
Δ
x
=
lim
Δ
x
→
0
2
x
+
2
Δ
x
+
3
−
2
x
−
3
)
Δ
x
=
lim
Δ
x
→
0
2
Δ
x
Δ
x
=
lim
Δ
x
→
0
2
=
2
{\displaystyle {\begin{aligned}f^{'}(x)&=\lim _{\Delta x\to 0}{\frac {(2(x+\Delta x)+3)-(2x+3)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x+2\Delta x+3-2x-3)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0}2\\&=\mathbf {2} \end{aligned}}}
3. 利用導數的定義,求函式
f
(
x
)
=
x
3
{\displaystyle f(x)=x^{3}}
的導數。現在嘗試
f
(
x
)
=
x
4
{\displaystyle f(x)=x^{4}}
。你能看出規律嗎?在下一節中,我們將找到
f
(
x
)
=
x
n
{\displaystyle f(x)=x^{n}}
對所有
n
{\displaystyle n}
的導數。
d
x
3
d
x
=
lim
Δ
x
→
0
(
x
+
Δ
x
)
3
−
x
3
Δ
x
d
x
4
d
x
=
lim
Δ
x
→
0
(
x
+
Δ
x
)
4
−
x
4
Δ
x
=
lim
Δ
x
→
0
x
3
+
3
x
2
Δ
x
+
3
x
Δ
x
2
+
Δ
x
3
−
x
3
Δ
x
=
lim
Δ
x
→
0
x
4
+
4
x
3
Δ
x
+
6
x
2
Δ
x
2
+
4
x
Δ
x
3
+
Δ
x
4
−
x
4
Δ
x
=
lim
Δ
x
→
0
3
x
2
Δ
x
+
3
x
Δ
x
2
+
Δ
x
3
Δ
x
=
lim
Δ
x
→
0
4
x
3
Δ
x
+
6
x
2
Δ
x
2
+
4
x
Δ
x
3
+
Δ
x
4
Δ
x
=
lim
Δ
x
→
0
3
x
2
+
3
x
Δ
x
+
Δ
x
2
=
lim
Δ
x
→
0
4
x
3
+
6
x
2
Δ
x
+
4
x
Δ
x
2
+
Δ
x
3
=
3
x
2
=
4
x
3
{\displaystyle {\begin{alignedat}{2}{\frac {dx^{3}}{dx}}&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{3}-x^{3}}{\Delta x}}&\qquad {\frac {dx^{4}}{dx}}&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{4}-x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {x^{3}+3x^{2}\Delta x+3x\Delta x^{2}+\Delta x^{3}-x^{3}}{\Delta x}}&&=\lim _{\Delta x\to 0}{\frac {x^{4}+4x^{3}\Delta x+6x^{2}\Delta x^{2}+4x\Delta x^{3}+\Delta x^{4}-x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {3x^{2}\Delta x+3x\Delta x^{2}+\Delta x^{3}}{\Delta x}}&&=\lim _{\Delta x\to 0}{\frac {4x^{3}\Delta x+6x^{2}\Delta x^{2}+4x\Delta x^{3}+\Delta x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}3x^{2}+3x\Delta x+\Delta x^{2}&&=\lim _{\Delta x\to 0}4x^{3}+6x^{2}\Delta x+4x\Delta x^{2}+\Delta x^{3}\\&=\mathbf {3x^{2}} &&=\mathbf {4x^{3}} \end{alignedat}}}
d
x
3
d
x
=
lim
Δ
x
→
0
(
x
+
Δ
x
)
3
−
x
3
Δ
x
d
x
4
d
x
=
lim
Δ
x
→
0
(
x
+
Δ
x
)
4
−
x
4
Δ
x
=
lim
Δ
x
→
0
x
3
+
3
x
2
Δ
x
+
3
x
Δ
x
2
+
Δ
x
3
−
x
3
Δ
x
=
lim
Δ
x
→
0
x
4
+
4
x
3
Δ
x
+
6
x
2
Δ
x
2
+
4
x
Δ
x
3
+
Δ
x
4
−
x
4
Δ
x
=
lim
Δ
x
→
0
3
x
2
Δ
x
+
3
x
Δ
x
2
+
Δ
x
3
Δ
x
=
lim
Δ
x
→
0
4
x
3
Δ
x
+
6
x
2
Δ
x
2
+
4
x
Δ
x
3
+
Δ
x
4
Δ
x
=
lim
Δ
x
→
0
3
x
2
+
3
x
Δ
x
+
Δ
x
2
=
lim
Δ
x
→
0
4
x
3
+
6
x
2
Δ
x
+
4
x
Δ
x
2
+
Δ
x
3
=
3
x
2
=
4
x
3
{\displaystyle {\begin{alignedat}{2}{\frac {dx^{3}}{dx}}&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{3}-x^{3}}{\Delta x}}&\qquad {\frac {dx^{4}}{dx}}&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{4}-x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {x^{3}+3x^{2}\Delta x+3x\Delta x^{2}+\Delta x^{3}-x^{3}}{\Delta x}}&&=\lim _{\Delta x\to 0}{\frac {x^{4}+4x^{3}\Delta x+6x^{2}\Delta x^{2}+4x\Delta x^{3}+\Delta x^{4}-x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {3x^{2}\Delta x+3x\Delta x^{2}+\Delta x^{3}}{\Delta x}}&&=\lim _{\Delta x\to 0}{\frac {4x^{3}\Delta x+6x^{2}\Delta x^{2}+4x\Delta x^{3}+\Delta x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}3x^{2}+3x\Delta x+\Delta x^{2}&&=\lim _{\Delta x\to 0}4x^{3}+6x^{2}\Delta x+4x\Delta x^{2}+\Delta x^{3}\\&=\mathbf {3x^{2}} &&=\mathbf {4x^{3}} \end{alignedat}}}
4. 文字指出
|
x
|
{\displaystyle \left|x\right|}
在
x
=
0
{\displaystyle x=0}
處沒有定義。利用導數的定義證明這一點。
lim
Δ
x
→
0
−
|
0
+
Δ
x
|
−
|
0
|
Δ
x
=
lim
Δ
x
→
0
−
−
Δ
x
Δ
x
lim
Δ
x
→
0
+
|
0
+
Δ
x
|
−
|
0
|
Δ
x
=
lim
Δ
x
→
0
+
Δ
x
Δ
x
=
lim
Δ
x
→
0
−
−
1
=
lim
Δ
x
→
0
+
1
=
−
1
=
1
{\displaystyle {\begin{alignedat}{2}\lim _{\Delta x\to 0^{-}}{\frac {\left|0+\Delta x\right|-\left|0\right|}{\Delta x}}&=\lim _{\Delta x\to 0^{-}}{\frac {-\Delta x}{\Delta x}}&\qquad \lim _{\Delta x\to 0^{+}}{\frac {\left|0+\Delta x\right|-\left|0\right|}{\Delta x}}&=\lim _{\Delta x\to 0^{+}}{\frac {\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0^{-}}-1&&=\lim _{\Delta x\to 0^{+}}1\\&=-1&&=1\end{alignedat}}}
由於在
x
=
0
{\displaystyle x=0}
處左極限和右極限不相等,因此極限不存在,所以
|
x
|
{\displaystyle \left|x\right|}
在
x
=
0
{\displaystyle x=0}
處不可導。
lim
Δ
x
→
0
−
|
0
+
Δ
x
|
−
|
0
|
Δ
x
=
lim
Δ
x
→
0
−
−
Δ
x
Δ
x
lim
Δ
x
→
0
+
|
0
+
Δ
x
|
−
|
0
|
Δ
x
=
lim
Δ
x
→
0
+
Δ
x
Δ
x
=
lim
Δ
x
→
0
−
−
1
=
lim
Δ
x
→
0
+
1
=
−
1
=
1
{\displaystyle {\begin{alignedat}{2}\lim _{\Delta x\to 0^{-}}{\frac {\left|0+\Delta x\right|-\left|0\right|}{\Delta x}}&=\lim _{\Delta x\to 0^{-}}{\frac {-\Delta x}{\Delta x}}&\qquad \lim _{\Delta x\to 0^{+}}{\frac {\left|0+\Delta x\right|-\left|0\right|}{\Delta x}}&=\lim _{\Delta x\to 0^{+}}{\frac {\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0^{-}}-1&&=\lim _{\Delta x\to 0^{+}}1\\&=-1&&=1\end{alignedat}}}
由於在
x
=
0
{\displaystyle x=0}
處左極限和右極限不相等,因此極限不存在,所以
|
x
|
{\displaystyle \left|x\right|}
在
x
=
0
{\displaystyle x=0}
處不可導。
6. 利用導數的定義證明
sin
x
{\displaystyle \sin x}
的導數是
cos
x
{\displaystyle \cos x}
。提示:使用合適的積化和差公式,以及
lim
t
→
0
sin
(
t
)
t
=
1
{\displaystyle \lim _{t\to 0}{\frac {\sin(t)}{t}}=1}
和
lim
t
→
0
cos
(
t
)
−
1
t
=
0
{\displaystyle \lim _{t\to 0}{\frac {\cos(t)-1}{t}}=0}
。
lim
Δ
x
→
0
sin
(
x
+
Δ
x
)
−
sin
(
x
)
Δ
x
=
lim
Δ
x
→
0
(
sin
(
x
)
cos
(
Δ
x
)
+
cos
(
x
)
sin
(
Δ
x
)
)
−
sin
(
x
)
Δ
x
=
lim
Δ
x
→
0
sin
(
x
)
(
cos
(
Δ
x
)
−
1
)
+
cos
(
x
)
sin
(
Δ
x
)
Δ
x
=
sin
(
x
)
⋅
lim
Δ
x
→
0
cos
(
Δ
x
)
−
1
Δ
x
+
cos
(
x
)
⋅
lim
Δ
x
→
0
sin
(
Δ
x
)
Δ
x
=
sin
(
x
)
⋅
0
+
cos
(
x
)
⋅
1
=
cos
(
x
)
{\displaystyle {\begin{aligned}\lim _{\Delta x\to 0}{\frac {\sin(x+\Delta x)-\sin(x)}{\Delta x}}&=\lim _{\Delta x\to 0}{\frac {(\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x))-\sin(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\sin(x)(\cos(\Delta x)-1)+\cos(x)\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\cdot \lim _{\Delta x\to 0}{\frac {\cos(\Delta x)-1}{\Delta x}}+\cos(x)\cdot \lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\cdot 0+\cos(x)\cdot 1\\&=\cos(x)\end{aligned}}}
lim
Δ
x
→
0
sin
(
x
+
Δ
x
)
−
sin
(
x
)
Δ
x
=
lim
Δ
x
→
0
(
sin
(
x
)
cos
(
Δ
x
)
+
cos
(
x
)
sin
(
Δ
x
)
)
−
sin
(
x
)
Δ
x
=
lim
Δ
x
→
0
sin
(
x
)
(
cos
(
Δ
x
)
−
1
)
+
cos
(
x
)
sin
(
Δ
x
)
Δ
x
=
sin
(
x
)
⋅
lim
Δ
x
→
0
cos
(
Δ
x
)
−
1
Δ
x
+
cos
(
x
)
⋅
lim
Δ
x
→
0
sin
(
Δ
x
)
Δ
x
=
sin
(
x
)
⋅
0
+
cos
(
x
)
⋅
1
=
cos
(
x
)
{\displaystyle {\begin{aligned}\lim _{\Delta x\to 0}{\frac {\sin(x+\Delta x)-\sin(x)}{\Delta x}}&=\lim _{\Delta x\to 0}{\frac {(\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x))-\sin(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\sin(x)(\cos(\Delta x)-1)+\cos(x)\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\cdot \lim _{\Delta x\to 0}{\frac {\cos(\Delta x)-1}{\Delta x}}+\cos(x)\cdot \lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\cdot 0+\cos(x)\cdot 1\\&=\cos(x)\end{aligned}}}
求解以下方程的導數
7.
f
(
x
)
=
42
{\displaystyle f(x)=42}
f
′
(
x
)
=
0
{\displaystyle \mathbf {f'(x)=0} }
f
′
(
x
)
=
0
{\displaystyle \mathbf {f'(x)=0} }
8.
f
(
x
)
=
6
x
+
10
{\displaystyle f(x)=6x+10}
f
′
(
x
)
=
6
{\displaystyle \mathbf {f'(x)=6} }
f
′
(
x
)
=
6
{\displaystyle \mathbf {f'(x)=6} }
9.
f
(
x
)
=
2
x
2
+
12
x
+
3
{\displaystyle f(x)=2x^{2}+12x+3}
f
′
(
x
)
=
4
x
+
12
{\displaystyle \mathbf {f'(x)=4x+12} }
f
′
(
x
)
=
4
x
+
12
{\displaystyle \mathbf {f'(x)=4x+12} }
分類
:
書籍:微積分
華夏公益教科書