【数学分析笔记】第4章第2节 导数的意义和性质（1）
4. 微分
4.2 导数的意义与性质
4.2.1 导数在物理中的背景
物体在OS方向上运动，位移函数为
s
=
s
(
t
)
s=s(t)
s=s(t)，求时刻
t
t
t的瞬时速度，找一个区间
[
t
,
t
+
△
t
]
[t,t+\bigtriangleup t]
[t,t+△t]，从时刻
t
t
t变到时刻
t
+
△
t
t+\bigtriangleup t
t+△t，则
△
s
=
s
(
t
+
△
t
)
−
s
(
t
)
\bigtriangleup s=s(t+\bigtriangleup t)-s(t)
△s=s(t+△t)−s(t)，
△
s
△
t
\frac{\bigtriangleup s}{\bigtriangleup t}
△t△s是这段时间的平均速度，而这段时间的瞬时速度为
v
(
t
)
=
lim
△
t
→
0
△
s
△
t
=
lim
△
t
→
0
s
(
t
+
△
t
)
−
s
(
t
)
△
t
=
s
′
(
t
)
v(t)=\lim\limits_{\bigtriangleup t\to 0}\frac{\bigtriangleup s}{\bigtriangleup t}=\lim\limits_{\bigtriangleup t\to 0}\frac{s(t+\bigtriangleup t)-s(t)}{\bigtriangleup t}=s'(t)
v(t)=△t→0lim△t△s=△t→0lim△ts(t+△t)−s(t)=s′(t)
再看一个例子，设
p
(
t
)
p(t)
p(t)表示某地区在
t
t
t时刻的人口数，要算某一时段的人口增长率，在
[
t
,
t
+
△
t
]
[t,t+\bigtriangleup t]
[t,t+△t]这段时间，
△
p
=
p
(
t
+
△
t
)
−
p
(
t
)
\bigtriangleup p=p(t+\bigtriangleup t)-p(t)
△p=p(t+△t)−p(t)，则
△
p
△
t
\frac{\bigtriangleup p}{\bigtriangleup t}
△t△p是这段时间人口平均增长率，人口在该时刻
t
t
t的增长率为
lim
△
t
→
0
△
p
△
t
=
lim
△
t
→
0
p
(
t
+
△
t
)
−
p
(
t
)
△
t
=
p
′
(
t
)
\lim\limits_{\bigtriangleup t\to 0}\frac{\bigtriangleup p}{\bigtriangleup t}=\lim\limits_{\bigtriangleup t\to 0}\frac{p(t+\bigtriangleup t)-p(t)}{\bigtriangleup t}=p'(t)
△t→0lim△t△p=△t→0lim△tp(t+△t)−p(t)=p′(t)
4.2.2 导数的几何意义
画一条曲线，求曲线上一点的切线的斜率
切线看成是割线的极限位置
让
△
x
→
0
\bigtriangleup x\to 0
△x→0，则割线的斜率为
△
y
△
x
\frac{\bigtriangleup y}{\bigtriangleup x}
△x△y，记切线斜率为
k
k
k，记函数为
y
=
f
(
x
)
y=f(x)
y=f(x)，切线斜率为
k
=
lim
△
x
→
0
△
y
△
x
=
lim
△
x
→
0
f
(
x
+
△
x
)
−
f
(
x
)
△
x
=
f
′
(
x
)
k=\lim\limits_{\bigtriangleup x\to 0}\frac{\bigtriangleup y}{\bigtriangleup x}=\lim\limits_{\bigtriangleup x\to 0}\frac{f(x+\bigtriangleup x)-f(x)}{\bigtriangleup x}=f'(x)
k=△x→0lim△x△y=△x→0lim△xf(x+△x)−f(x)=f′(x)
过
(
x
0
,
f
(
x
0
)
)
(x_{0},f(x_{0}))
(x0,f(x0))的切线：
y
−
f
(
x
0
)
=
f
′
(
x
0
)
(
x
−
x
0
)
y-f(x_0)=f'(x_0)(x-x_0)
y−f(x0)=f′(x0)(x−x0)，法线方程为
y
−
f
(
x
0
)
=
−
1
f
′
(
x
0
)
(
x
−
x
0
)
,
f
′
(
x
0
)
≠
0
y-f(x_0)=-\frac{1}{f'(x_0)}(x-x_0),f'(x_0)\ne 0
y−f(x0)=−f′(x0)1(x−x0),f′(x0)=0
【例4.2.1】求抛物线
y
2
=
2
p
x
,
p
>
0
y^2=2px,p>0
y2=2px,p>0，
(
x
0
,
y
0
)
(x_0,y_0)
(x0,y0)是抛物线上切线一点，求过
(
x
0
,
y
0
)
(x_0,y_0)
(x0,y0)的切线方程。
【解】
y
=
2
p
x
y=\sqrt{2px}
y=2px
k
=
lim
△
x
→
0
2
p
(
x
+
△
x
)
−
2
p
x
△
x
=
lim
△
x
→
0
(
2
p
(
x
+
△
x
)
−
2
p
x
)
(
2
p
(
x
+
△
x
)
+
2
p
x
)
△
x
(
2
p
(
x
+
△
x
)
+
2
p
x
)
=
lim
△
x
→
0
2
p
△
x
△
x
(
2
p
(
x
+
△
x
)
+
2
p
x
)
=
lim
△
x
→
0
2
p
2
p
(
x
+
△
x
)
+
2
p
x
=
2
p
2
2
p
x
=
p
2
x
k=\lim\limits_{\bigtriangleup x\to 0}\frac{\sqrt{2p(x+\bigtriangleup x)}-\sqrt{2px}}{\bigtriangleup x}=\lim\limits_{\bigtriangleup x\to 0}\frac{(\sqrt{2p(x+\bigtriangleup x)}-\sqrt{2px})(\sqrt{2p(x+\bigtriangleup x)}+\sqrt{2px})}{\bigtriangleup x(\sqrt{2p(x+\bigtriangleup x)}+\sqrt{2px})}=\lim\limits_{\bigtriangleup x\to 0}\frac{2p\bigtriangleup x}{\bigtriangleup x(\sqrt{2p(x+\bigtriangleup x)}+\sqrt{2px})}=\lim\limits_{\bigtriangleup x\to 0}\frac{2p}{\sqrt{2p(x+\bigtriangleup x)}+\sqrt{2px}}=\frac{2p}{2\sqrt{2px}}=\sqrt{\frac{p}{2x}}
k=△x→0lim△x2p(x+△x)−2px=△x→0lim△x(2p(x+△x)+2px)(2p(x+△x)−2px)(2p(x+△x)+2px)=△x→0lim△x(2p(x+△x)+2px)2p△x=△x→0lim2p(x+△x)+2px2p=22px2p=2xp
切线方程为
y
−
y
0
=
p
2
x
0
(
x
−
x
0
)
y-y_0=\sqrt{\frac{p}{2x_0}}(x-x_0)
y−y0=2x0p(x−x0)
【注】由抛物线的切线方程得到抛物线的重要性质：
tan
θ
1
=
p
2
x
0
=
p
y
0
\tan \theta_{1}=\frac{\sqrt{p}}{\sqrt{2 x_{0}}}=\frac{p}{y_{0}}
tanθ1=2x0p=y0p，记
(
x
0
,
y
0
)
(x_0,y_0)
(x0,y0)与抛物线焦点
(
p
2
,
0
)
(\frac{p}{2},0)
(2p,0)连线与
x
x
x轴的夹角为
θ
2
\theta_{2}
θ2，该连线与抛物线在点
(
x
0
,
y
0
)
(x_0,y_0)
(x0,y0)处的切线的夹角为
θ
\theta
θ，则有：
tan
θ
2
=
y
0
x
0
−
p
2
\tan \theta_{2}=\frac{y_{0}}{x_{0}-\frac{p}{2}}
tanθ2=x0−2py0
于是
tan
θ
=
tan
(
θ
2
−
θ
1
)
=
tan
θ
2
−
tan
θ
1
1
+
tan
θ
2
⋅
tan
θ
1
=
y
0
x
0
−
p
2
−
p
y
0
1
+
y
0
x
0
−
p
2
⋅
p
y
0
=
p
y
0
=
tan
θ
1
\tan \theta=\tan (\theta_{2} - \theta_{1})=\frac{\tan \theta_{2}-\tan \theta_{1}}{1+\tan \theta_{2} \cdot \tan \theta_{1}}=\frac{\frac{y_{0}}{x_{0}-\frac{p}{2}}-\frac{p}{y_{0}}}{1+\frac{y_{0}}{x_{0}-\frac{p}{2}} \cdot \frac{p}{y_{0}}}=\frac{p}{y_{0}}=\tan \theta_{1}
tanθ=tan(θ2−θ1)=1+tanθ2⋅tanθ1tanθ2−tanθ1=1+x0−2py0⋅y0px0−2py0−y0p=y0p=tanθ1（
y
0
2
=
2
p
x
0
y_{0}^{2}=2px_{0}
y02=2px0代入化简计算）
即恰好
θ
\theta
θ与切线与
x
x
x轴夹角
θ
1
\theta_{1}
θ1相等。
【例4.2.2】
x
2
a
2
+
y
2
b
2
=
1
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
a2x2+b2y2=1，求椭圆上过
(
x
0
,
y
0
)
(x_0,y_0)
(x0,y0)点的切线。
【解】不妨设
y
0
>
0
y_0>0
y0>0，
y
=
b
a
a
2
−
x
2
y=\frac{b}{a}\sqrt{a^2-x^2}
y=aba2−x2
k
=
b
a
lim
Δ
x
→
0
a
2
−
(
x
0
+
Δ
x
)
2
−
a
2
−
x
0
2
Δ
x
=
b
a
lim
Δ
x
→
0
x
0
2
−
(
x
0
+
Δ
x
)
2
(
a
2
−
(
x
0
+
Δ
x
)
2
+
a
2
−
x
0
2
)
⋅
Δ
x
=
b
a
−
x
0
a
2
−
x
0
2
k=\frac{b}{a} \lim\limits_{\Delta x \rightarrow 0} \frac{\sqrt{a^{2}-\left(x_{0}+\Delta x\right)^{2}}-\sqrt{a^{2}-x_{0}^{2}}}{\Delta x}=\frac{b}{a} \lim\limits_{\Delta x \rightarrow 0} \frac{x_{0}^{2}-\left(x_{0}+\Delta x\right)^{2}}{\left(\sqrt{a^{2}-\left(x_{0}+\Delta x\right)^{2}}+\sqrt{a^{2}-x_{0}^{2}}\right) \cdot \Delta x}=\frac{b}{a} \frac{-x_{0}}{\sqrt{a^{2}-x_{0}^{2}}}
k=abΔx→0limΔxa2−(x0+Δx)2−a2−x02=abΔx→0lim(a2−(x0+Δx)2+a2−x02)⋅Δxx02−(x0+Δx)2=aba2−x02−x0
所以切线方程为
y
−
y
0
=
b
a
−
x
0
a
2
−
x
0
2
(
x
−
x
0
)
=
−
b
2
a
2
⋅
x
0
b
a
a
2
−
x
0
2
y-y_{0}=\frac{b}{a} \frac{-x_{0}}{\sqrt{a^{2}-x_{0}^{2}}}\left(x-x_{0}\right)=-\frac{b^2}{a^2}\cdot\frac{x_0}{\frac{b}{a}\sqrt{a^2-x_{0}^{2}}}
y−y0=aba2−x02−x0(x−x0)=−a2b2⋅aba2−x02x0
即
a
2
y
0
y
+
b
2
x
0
x
=
a
2
y
0
2
+
b
2
x
0
2
=
a
2
b
2
a^2y_0y+b^2x_0x=a^2y_0^2+b^2x_0^2=a^2b^2
a2y0y+b2x0x=a2y02+b2x02=a2b2
即
y
0
y
a
+
x
0
x
b
=
1
\frac{y_0y}{a}+\frac{x_0x}{b}=1
ay0y+bx0x=1
原文地址：https://blog.csdn.net/qq_30204431/article/details/142649532
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