背景
在学习高等数学的极限和导数时,发现老师给了很多等价无穷小公式以及求导法则,记忆力有限,死记硬背容易背叉,所以尝试动手推导其过程,加深理解。
推导原则:利用两个重要极限和无穷小替换推导,尽可能不用后面的知识,如洛必达法则,泰勒公式等。后面公式的推导可能会用到前面已经推导的结论(因为按照教程顺序还没学到后面知识~)。
常见无穷小
常见求导法则
两个重要极限
这两个重要极限不做推导,只有死记硬背,作为后续基础。
第一个重要极限
lim x → 0 x s i n x = 1 \lim_{x\rightarrow0}\frac{x}{sinx}=1 x→0limsinxx=1
第二个重要极限
lim
x
→
+
∞
(
1
+
1
x
)
x
=
e
\lim_{x\rightarrow+\infty}(1+\frac{1}{x})^x=e
x→+∞lim(1+x1)x=e
在我当前的理解里,后续的洛必达法则其实就是求导,对指数及对数函数来说,求导和无穷小替换都会用到这个重要极限,这也是为什么我在之前的文章说:在证明这个重要极限时不能使用洛必达法则
等价无穷小
定义,当 f ( x ) → 0 , g ( x ) → 0 f(x)\rightarrow0,g(x)\rightarrow0 f(x)→0,g(x)→0时,如果 lim x → 0 f ( x ) g ( x ) = 1 \lim_{x\rightarrow0}\frac{f(x)}{g(x)}=1 limx→0g(x)f(x)=1,那么说g(x)和f(x)为等价无穷小,记作f(x)~g(x)
tanx~x
lim x → 0 t a n x x = lim x → 0 s i n x c o s x ∗ x = lim x → 0 1 c o s x = 1 \lim_{x\rightarrow0}\frac{tanx}{x}=\lim_{x\rightarrow0}\frac{sinx}{cosx*x}=\lim_{x\rightarrow0}\frac{1}{cosx}=1 limx→0xtanx=limx→0cosx∗xsinx=limx→0cosx1=1
arcsinx~x
lim x → 0 a r c s i n x x = lim t → 0 t s i n t = 1 \lim_{x\rightarrow0}\frac{arcsinx}{x}=\lim_{t\rightarrow0}\frac{t}{sint}=1 limx→0xarcsinx=limt→0sintt=1
原函数不好理解,可以考虑换元法,即转为反函数。令arcsinx=t,则sint=x。后续不再单独解释
arctanx~x
lim x → 0 a r c t a n x x = lim t → 0 t t a n t = 1 \lim_{x\rightarrow0}\frac{arctanx}{x}=\lim_{t\rightarrow0}\frac{t}{tant}=1 limx→0xarctanx=limt→0tantt=1
ln(1+x)~x
lim x → 0 l n ( 1 + x ) x = lim x → 0 l n ( 1 + x ) 1 x = l n ( lim x → + ∞ ( 1 + 1 x ) x ) = l n e = 1 \lim_{x\rightarrow0}\frac{ln(1+x)}{x}= \lim_{x\rightarrow0}ln(1+x)^\frac{1}{x}= ln( \lim_{x\rightarrow+\infty}(1+\frac{1}{x})^x)= lne=1 x→0limxln(1+x)=x→0limln(1+x)x1=ln(x→+∞lim(1+x1)x)=lne=1
e^x-1~x
令
e
x
−
1
=
t
e^x-1=t
ex−1=t,则x=ln(1+t)。因为
x
→
0
x\rightarrow0
x→0,所以
t
→
0
t\rightarrow0
t→0
lim
x
→
0
e
x
−
1
x
=
lim
t
→
0
t
l
n
(
1
+
t
)
=
1
\lim_{x\rightarrow0}\frac{e^x-1}{x}=\lim_{t\rightarrow0}\frac{t}{ln(1+t)}=1
x→0limxex−1=t→0limln(1+t)t=1
a^x-1~xlna
预备知识
对数函数不好理解,转为指数函数就好了
假设
log
a
(
t
+
1
)
=
m
,
ln
a
=
n
\log_a(t+1)=m,\ln a=n
loga(t+1)=m,lna=n,那么有:
a
m
=
t
+
1
,
e
n
=
a
a^m=t+1,e^n=a
am=t+1,en=a,所以,
(
e
n
)
m
=
t
+
1
(e^n)^m=t+1
(en)m=t+1,即:
e
m
n
=
t
+
1
=
>
m
∗
n
=
l
n
(
t
+
1
)
e^{mn}=t+1=>m*n=ln(t+1)
emn=t+1=>m∗n=ln(t+1)
所以有结论:
log
a
(
t
+
1
)
∗
ln
a
=
m
∗
n
=
ln
(
t
+
1
)
\log_a(t+1)*\ln a=m*n=\ln(t+1)
loga(t+1)∗lna=m∗n=ln(t+1)
正式推导:
lim x → 0 a x − 1 x ln a = lim t → 0 t log a ( t + 1 ) ln a = lim t → 0 t ln ( t + 1 ) = 1 \lim_{x\rightarrow0}\frac{a^x-1}{x\ln a}=\lim_{t\rightarrow0}\frac{t}{\log_a(t+1)\ln a}=\lim_{t\rightarrow0}\frac{t}{\ln(t+1)}=1 x→0limxlnaax−1=t→0limloga(t+1)lnat=t→0limln(t+1)t=1
(1+bx)^a-1~abx
因为
lim
x
→
0
l
n
(
1
+
x
)
x
=
1
\lim_{x\rightarrow0}\frac{ln(1+x)}{x}=1
limx→0xln(1+x)=1
所以
lim
t
→
1
l
n
(
t
)
t
−
1
=
1
\lim_{t\rightarrow1}\frac{ln(t)}{t-1}=1
limt→1t−1ln(t)=1
所以下面令t=(1+bx)^a,
x
→
0
,
则
t
→
1
x\rightarrow0,则t\rightarrow1
x→0,则t→1。
所以利用等价无穷小
lim
t
→
1
l
n
(
t
)
t
−
1
=
1
\lim_{t\rightarrow1}\frac{ln(t)}{t-1}=1
limt→1t−1ln(t)=1 替换:
lim
x
→
0
(
1
+
b
x
)
a
−
1
a
b
x
=
lim
x
→
0
l
n
(
1
+
b
x
)
a
a
b
x
=
lim
x
→
0
l
n
(
1
+
b
x
)
b
x
=
1
\lim_{x\rightarrow0}\frac{(1+bx)^a-1}{abx}=\lim_{x\rightarrow0}\frac{ln(1+bx)^a}{abx}=\lim_{x\rightarrow0}\frac{ln(1+bx)}{bx}=1
limx→0abx(1+bx)a−1=limx→0abxln(1+bx)a=limx→0bxln(1+bx)=1
(1+x)^(1/n)-1 ~ x/n
1
+
x
n
−
1
\sqrt[n]{1+x}-1
n1+x−1 ~
1
n
x
\frac{1}{n}x
n1x
预备知识:
a
n
−
1
=
a
n
+
a
n
−
1
+
a
n
−
2
+
.
.
.
+
a
−
(
a
n
−
1
+
a
n
−
2
+
.
.
.
+
a
)
−
1
a^n-1=a^n+a^{n-1}+a^{n-2}+...+a-(a^{n-1}+a^{n-2}+...+a)-1
an−1=an+an−1+an−2+...+a−(an−1+an−2+...+a)−1
=
a
n
+
a
n
−
1
+
a
n
−
2
+
.
.
.
+
a
−
(
a
n
−
1
+
a
n
−
2
+
.
.
.
+
a
+
1
)
=a^n+a^{n-1}+a^{n-2}+...+a-(a^{n-1}+a^{n-2}+...+a+1)
=an+an−1+an−2+...+a−(an−1+an−2+...+a+1)
=
a
∗
(
a
n
−
1
+
a
n
−
2
+
.
.
.
+
a
+
1
)
−
(
a
n
−
1
+
a
n
−
2
+
.
.
.
+
a
+
1
)
=a*(a^{n-1}+a^{n-2}+...+a+1)-(a^{n-1}+a^{n-2}+...+a+1)
=a∗(an−1+an−2+...+a+1)−(an−1+an−2+...+a+1)
=
(
a
−
1
)
(
a
n
−
1
+
a
n
−
2
+
.
.
.
+
a
+
1
)
=(a-1)(a^{n-1}+a^{n-2}+...+a+1)
=(a−1)(an−1+an−2+...+a+1)
正式推导
lim
x
→
0
1
+
x
n
−
1
x
n
\lim_{x\rightarrow0}\frac{\sqrt[n]{1+x}-1}{\frac{x}{n}}
limx→0nxn1+x−1
=
lim
x
→
0
(
1
+
x
)
1
n
−
1
x
n
=\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{n}}-1}{\frac{x}{n}}
=limx→0nx(1+x)n1−1
将
(
1
+
x
)
1
n
(1+x)^{\frac{1}{n}}
(1+x)n1看作预备知识中的a,则
原式=
lim
x
→
0
n
∗
(
1
+
x
−
1
)
x
∗
(
(
1
+
x
)
n
−
1
n
+
(
1
+
x
)
n
−
2
n
+
.
.
.
+
(
1
+
x
)
1
n
+
1
)
\lim_{x\rightarrow0}\frac{n*(1+x-1)}{x*((1+x)^{\frac{n-1}{n}}+(1+x)^{\frac{n-2}{n}}+...+(1+x)^{\frac{1}{n}}+1 )}
limx→0x∗((1+x)nn−1+(1+x)nn−2+...+(1+x)n1+1)n∗(1+x−1)
= lim x → 0 n x x ∗ ( n − 1 + 1 ) = 1 =\lim_{x\rightarrow0}\frac{nx}{x*(n-1+1)}=1 =limx→0x∗(n−1+1)nx=1
loga(1+x)~ x/lna
lim x → 0 log a ( 1 + x ) x ln a = lim x → 0 ln a ∗ log a ( 1 + x ) x = lim x → 0 l n ( 1 + x ) x = 1 \lim_{x\rightarrow0}\frac{\log_a(1+x)}{\frac{x}{\ln a}}=\lim_{x\rightarrow0}\frac{\ln a*\log_a(1+x)}{x}=\lim_{x\rightarrow0}\frac{ln(1+x)}{x}=1 limx→0lnaxloga(1+x)=limx→0xlna∗loga(1+x)=limx→0xln(1+x)=1
tanx-sinx~ 1/2x^3
lim x → 0 t a n x − s i n x 1 2 ∗ x 3 = lim x → 0 s i n x ∗ ( 1 − c o s x ) c o s x 1 2 ∗ x 3 = lim x → 0 2 ∗ s i n x ∗ ( 1 − c o s x ) x 3 ∗ c o s x = lim x → 0 2 ∗ ( 1 − c o s x ) sin 2 x ∗ c o s x = lim x → 0 2 ∗ ( 1 − c o s x ) ( 1 + c o s x ) ∗ ( 1 − c o s x ) ∗ c o s x = lim x → 0 2 ( 1 + c o s x ) ∗ c o s x = 1 \lim_{x\rightarrow0}\frac{tanx-sinx}{\frac{1}{2}*x^3}=\lim_{x\rightarrow0}\frac{\frac{sinx*(1-cosx)}{cosx}}{\frac{1}{2}*x^3}=\lim_{x\rightarrow0}\frac{2*sinx*(1-cosx)}{x^3*cosx}=\lim_{x\rightarrow0}\frac{2*(1-cosx)}{\sin^2x*cosx}=\lim_{x\rightarrow0}\frac{2*(1-cosx)}{(1+cosx)*(1-cosx)*cosx}=\lim_{x\rightarrow0}\frac{2}{(1+cosx)*cosx}=1 limx→021∗x3tanx−sinx=limx→021∗x3cosxsinx∗(1−cosx)=limx→0x3∗cosx2∗sinx∗(1−cosx)=limx→0sin2x∗cosx2∗(1−cosx)=limx→0(1+cosx)∗(1−cosx)∗cosx2∗(1−cosx)=limx→0(1+cosx)∗cosx2=1
1-cosx ~ x^2/2
lim x → 0 1 − c o s x x 2 2 = lim x → 0 2 ∗ ( 2 ∗ sin 2 x 2 ) x 2 = lim x → 0 x 2 x 2 = 1 \lim_{x\rightarrow0}\frac{1-cosx}{\frac{x^2}{2}}=\lim_{x\rightarrow0}\frac{2*(2*\sin^2\frac{x}{2})}{x^2}=\lim_{x\rightarrow0}\frac{x^2}{x^2}=1 limx→02x21−cosx=limx→0x22∗(2∗sin22x)=limx→0x2x2=1
x-sinx~ 1/6x^3
暂时无法仅利用极限或等价无穷小知识证明,必须用到洛必达法则。
洛必达法则理解参考:https://blog.csdn.net/qq_31076523/article/details/104737300
在运用洛必达法则的过程中会用到求导知识,本例来说用到了以下三类,这三类公式的证明参考后面的求导法则证明
C
′
=
0
(
C
为
常
数
)
C'=0(C为常数)
C′=0(C为常数)
(
x
n
)
′
=
(
n
−
1
)
∗
x
(x^n)'=(n-1)*x
(xn)′=(n−1)∗x
s
i
n
′
(
x
)
=
c
o
s
x
sin'(x)=cosx
sin′(x)=cosx
c
o
s
′
(
x
)
=
−
s
i
n
x
cos'(x)=-sinx
cos′(x)=−sinx
lim x → 0 x − s i n x x 3 6 = lim x → 0 1 − c o s x 3 x 2 6 = lim x → 0 s i n x x = 1 \lim_{x\rightarrow0}\frac{x-sinx}{\frac{x^3}{6}}=\lim_{x\rightarrow0}\frac{1-cosx}{\frac{3x^2}{6}}=\lim_{x\rightarrow0}\frac{sinx}{x}=1 limx→06x3x−sinx=limx→063x21−cosx=limx→0xsinx=1
tanx-x~ 1/3x^3
运用洛必达法则时用到了求导法则 t a n x = 1 cos 2 x tanx=\frac{1}{\cos^2x} tanx=cos2x1
lim x → 0 t a n x − x x 3 3 = lim x → 0 1 c o s 2 x − 1 x 2 = lim x → 0 tan 2 x x 2 = 1 \lim_{x\rightarrow0}\frac{tanx-x}{\frac{x^3}{3}}=\lim_{x\rightarrow0}\frac{\frac{1}{cos^2x}-1}{x^2}=\lim_{x\rightarrow0}\frac{\tan^2x}{x^2}=1 limx→03x3tanx−x=limx→0x2cos2x1−1=limx→0x2tan2x=1
x-ln(1+x) ~ x^2/2
运用洛必达法则时用到了求导法则 l n ( 1 + x ) = 1 x + 1 ln(1+x)=\frac{1}{x+1} ln(1+x)=x+11
lim x → 0 x − l n ( 1 + x ) x 2 2 = lim x → 0 1 − 1 1 + x x = lim x → 0 x ( 1 + x ) ∗ x = lim x → 0 1 ( 1 + x ) = 1 \lim_{x\rightarrow0}\frac{x-ln(1+x)}{\frac{x^2}{2}}=\lim_{x\rightarrow0}\frac{1-\frac{1}{1+x}}{x}=\lim_{x\rightarrow0}\frac{x}{(1+x)*x}=\lim_{x\rightarrow0}\frac{1}{(1+x)}=1 limx→02x2x−ln(1+x)=limx→0x1−1+x1=limx→0(1+x)∗xx=limx→0(1+x)1=1
最后三例均通过洛必达法则证明,可以看到简化了很多计算。
求导法则
求导用定义证明就是求极限。
求极限时,形式为
0
0
\frac{0}{0}
00 型时,可以根据上述的等价无穷小进行替换。
(x^n)'=(n-1)*x
(
x
n
)
′
=
lim
t
→
0
(
x
+
t
)
n
−
x
n
t
=
lim
t
→
0
(
C
n
0
∗
x
n
+
C
n
1
∗
x
n
−
1
t
+
C
n
2
∗
x
n
−
2
∗
t
2
+
.
.
.
+
C
n
n
t
n
)
−
x
n
t
=
lim
t
→
0
(
C
n
1
∗
x
n
−
1
t
+
C
n
2
∗
x
n
−
2
∗
t
2
+
.
.
.
+
C
n
n
t
n
)
t
(x^n)'=\lim_{t\rightarrow 0}\frac{(x+t)^n-x^n}{t}=\lim_{t\rightarrow 0}\frac{(C_n^0*x^n+C_n^1*x^{n-1}t+C_n^2*x^{n-2}*t^2+...+C_n^nt^n)-x^n}{t} =\lim_{t\rightarrow 0}\frac{(C_n^1*x^{n-1}t+C_n^2*x^{n-2}*t^2+...+C_n^nt^n)}{t}
(xn)′=limt→0t(x+t)n−xn=limt→0t(Cn0∗xn+Cn1∗xn−1t+Cn2∗xn−2∗t2+...+Cnntn)−xn=limt→0t(Cn1∗xn−1t+Cn2∗xn−2∗t2+...+Cnntn)
=
lim
t
→
0
(
C
n
1
∗
x
n
−
1
+
C
n
2
∗
x
n
−
2
t
+
.
.
.
+
C
n
n
∗
t
n
−
1
=\lim_{t\rightarrow 0}(C_n^1*x^{n-1}+C_n^2*x^{n-2}t+...+C_n^n*t^{n-1}
=limt→0(Cn1∗xn−1+Cn2∗xn−2t+...+Cnn∗tn−1)
=
n
∗
x
n
−
1
=n*x^{n-1}
=n∗xn−1
sin’(x)=cosx
x
+
t
−
x
2
\frac{x+t-x}{2}
2x+t−x
预备知识:
s
i
n
a
+
s
i
n
b
=
s
i
n
(
a
+
b
2
+
a
−
b
2
)
+
s
i
n
(
a
+
b
2
−
a
−
b
2
)
sina+sinb=sin(\frac{a+b}{2}+\frac{a-b}{2})+sin(\frac{a+b}{2}-\frac{a-b}{2})
sina+sinb=sin(2a+b+2a−b)+sin(2a+b−2a−b)
=
s
i
n
(
a
+
b
2
)
∗
c
o
s
(
a
−
b
2
)
+
c
o
s
(
a
+
b
2
)
∗
s
i
n
(
a
−
b
2
)
+
s
i
n
(
a
+
b
2
)
∗
c
o
s
(
a
−
b
2
)
−
c
o
s
(
a
+
b
2
)
∗
s
i
n
(
a
−
b
2
)
=sin(\frac{a+b}{2})*cos(\frac{a-b}{2})+cos(\frac{a+b}{2})*sin(\frac{a-b}{2})+sin(\frac{a+b} {2})*cos(\frac{a-b}{2})-cos(\frac{a+b}{2})*sin(\frac{a-b}{2})
=sin(2a+b)∗cos(2a−b)+cos(2a+b)∗sin(2a−b)+sin(2a+b)∗cos(2a−b)−cos(2a+b)∗sin(2a−b)
=
2
s
i
n
(
a
+
b
2
)
∗
c
o
s
(
a
−
b
2
)
=2sin(\frac{a+b}{2})*cos(\frac{a-b}{2})
=2sin(2a+b)∗cos(2a−b)
求导推导:
s
i
n
′
x
=
lim
t
→
0
s
i
n
(
x
+
t
)
−
s
i
n
(
x
)
t
=
lim
t
→
0
s
i
n
(
x
+
t
)
+
s
i
n
(
−
x
)
t
sin'x=\lim_{t\rightarrow 0}\frac{sin(x+t)-sin(x)}{t} =\lim_{t\rightarrow 0}\frac{sin(x+t)+sin(-x)}{t}
sin′x=limt→0tsin(x+t)−sin(x)=limt→0tsin(x+t)+sin(−x)
= lim t → 0 2 s i n ( x + t − x 2 ) ∗ c o s ( x + t + x 2 ) t = lim t → 0 2 s i n t 2 c o s 2 x + t 2 t =\lim_{t\rightarrow 0}\frac{2sin(\frac{x+t-x}{2})*cos(\frac{x+t+x}{2})}{t} =\lim_{t\rightarrow 0}\frac{2sin\frac{t}{2}cos\frac{2x+t}{2}}{t} =limt→0t2sin(2x+t−x)∗cos(2x+t+x)=limt→0t2sin2tcos22x+t
= lim t → 0 t ∗ c o s ( x + t 2 ) t = c o s x =\lim_{t\rightarrow 0}\frac{t*cos(x+\frac{t}{2})}{t}=cosx =limt→0tt∗cos(x+2t)=cosx
cos’(x)=-sinx
方法一:
c
o
s
′
x
=
s
i
n
′
(
π
2
−
x
)
=
c
o
s
(
π
2
−
x
)
∗
−
1
=
−
s
i
n
x
cos'x=sin'(\frac{\pi}{2}-x)=cos(\frac{\pi}{2}-x)*-1=-sinx
cos′x=sin′(2π−x)=cos(2π−x)∗−1=−sinx
方法二:
c
o
s
′
x
=
(
1
−
2
s
i
n
2
x
2
)
′
=
−
4
∗
s
i
n
(
x
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cos'x=(1-2sin^2\frac{x}{2})'=-4*sin(\frac{x}{2})*cos(\frac{x}{2})*\frac{1}{2}=-2*sin(\frac{x} {2})*cos(\frac{x}{2})=-sinx
cos′x=(1−2sin22x)′=−4∗sin(2x)∗cos(2x)∗21=−2∗sin(2x)∗cos(2x)=−sinx
tan’x=1/cos^2x
t a n x = ( s i n x c o s x ) ′ = c o s 2 x + s i n 2 x cos 2 x = 1 c o s 2 x = s e c 2 x tanx=(\frac{sinx}{cosx})'=\frac{cos^2x+sin^2x}{\cos^2x}=\frac{1}{cos^2x}=sec^2x tanx=(cosxsinx)′=cos2xcos2x+sin2x=cos2x1=sec2x
ln’(1+x)=1/x+1
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ln'(1+x)=\lim_{t\rightarrow 0}\frac{ln(1+x+t)-ln(1+x)}{t}=\lim_{t\rightarrow 0}\frac{1}{t}ln(\frac{1+x+t}{1+x})
ln′(1+x)=limt→0tln(1+x+t)−ln(1+x)=limt→0t1ln(1+x1+x+t)
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=\lim_{t\rightarrow 0}ln(1+\frac{t}{1+x})^{\frac{1+x}{t}*\frac{1}{1+x}}=lne^{\frac{1}{1+x}}=\frac{1}{1+x}
=limt→0ln(1+1+xt)t1+x∗1+x1=lne1+x1=1+x1
(e^x)'= e^x
( e x ) ′ = lim t → 0 e x + t − e x t = lim t → 0 e x ( e t − 1 ) t = e x (e^x)'=\lim_{t\rightarrow 0}\frac{e^{x+t}-e^x}{t}=\lim_{t\rightarrow 0}\frac{e^x(e^t-1)}{t}=e^x (ex)′=t→0limtex+t−ex=t→0limtex(et−1)=ex文章来源:https://www.toymoban.com/news/detail-402074.html
(a^x)'= a^xlna
( a x ) ′ = lim t → 0 a x + t − a x t = lim t → 0 a x ( a t − 1 ) t = lim t → 0 a x l n a ∗ t t = a x l n a (a^x)'=\lim_{t\rightarrow 0}\frac{a^{x+t}-a^x}{t}=\lim_{t\rightarrow 0}\frac{a^x(a^t-1)}{t}=\lim_{t\rightarrow 0}\frac{a^xlna*t}{t}=a^xlna (ax)′=t→0limtax+t−ax=t→0limtax(at−1)=t→0limtaxlna∗t=axlna文章来源地址https://www.toymoban.com/news/detail-402074.html
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