【数学分析笔记】第4章第3节 导数四则运算和反函数求导法则（1）
4. 微分
4.3 导数四则运算与反函数求导法则
通过例题，计算常用的基本初等函数的导数
【例4.3.1】
y
=
sin
x
y=\sin x
y=sinx
【解】
y
′
(
x
)
=
lim
Δ
x
→
0
sin
(
x
+
Δ
x
)
−
sin
x
Δ
x
=
lim
Δ
x
→
0
2
cos
2
x
+
Δ
x
2
sin
Δ
x
2
Δ
x
=
cos
x
y'(x)=\lim\limits_{\Delta x\to 0}\frac{\sin (x+\Delta x)-\sin x}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{2\cos\frac{2x+\Delta x}{2}\sin\frac{\Delta x}{2}}{\Delta x}= \cos x
y′(x)=Δx→0limΔxsin(x+Δx)−sinx=Δx→0limΔx2cos22x+Δxsin2Δx=cosx
同理
y
=
cos
x
,
y
′
=
−
sin
x
y=\cos x,y' = -\sin x
y=cosx,y′=−sinx
【例4.3.2】
y
=
ln
x
y=\ln x
y=lnx
【解】
y
′
(
x
)
=
lim
Δ
x
→
0
ln
(
x
+
Δ
x
)
−
ln
x
Δ
x
=
lim
Δ
x
→
0
ln
(
1
+
Δ
x
x
)
Δ
x
=
lim
Δ
x
→
0
Δ
x
x
Δ
x
=
1
x
y'(x)=\lim\limits_{\Delta x\to 0}\frac{\ln(x+\Delta x)-\ln x}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{\ln(1+\frac{\Delta x}{x})}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{\frac{\Delta x}{x}}{\Delta x}=\frac{1}{x}
y′(x)=Δx→0limΔxln(x+Δx)−lnx=Δx→0limΔxln(1+xΔx)=Δx→0limΔxxΔx=x1
【例4.3.3】
y
=
e
x
y=e^x
y=ex
【解】
y
′
(
x
)
=
lim
Δ
x
→
0
e
x
+
Δ
x
−
e
x
Δ
x
=
lim
Δ
x
→
0
e
x
(
e
Δ
x
−
1
)
Δ
x
=
lim
Δ
x
→
0
e
x
Δ
x
Δ
x
=
e
x
y'(x)=\lim\limits_{\Delta x\to 0}\frac{e^{x+\Delta x}-e^x}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{e^x(e^{\Delta x}-1)}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{e^x\Delta x}{\Delta x}=e^x
y′(x)=Δx→0limΔxex+Δx−ex=Δx→0limΔxex(eΔx−1)=Δx→0limΔxexΔx=ex
【注】
y
=
a
x
=
e
ln
a
x
=
e
x
ln
a
,
y
′
(
x
)
=
lim
Δ
x
→
0
e
(
x
+
Δ
x
)
ln
a
−
e
x
ln
a
Δ
x
=
lim
Δ
x
→
0
e
x
ln
a
(
e
Δ
x
ln
a
−
1
)
Δ
x
=
lim
Δ
x
→
0
e
x
ln
a
Δ
x
ln
a
Δ
x
=
e
x
ln
a
ln
a
=
a
x
ln
a
y=a^x=e^{\ln a^x}=e^{x\ln a},y'(x)=\lim\limits_{\Delta x\to 0}\frac{e^{(x+\Delta x)\ln a}-e^{x\ln a}}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{e^{x\ln a}(e^{\Delta x\ln a}-1)}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{e^{x\ln a}\Delta x\ln a}{\Delta x}=e^{x\ln a}\ln a=a^x\ln a
y=ax=elnax=exlna,y′(x)=Δx→0limΔxe(x+Δx)lna−exlna=Δx→0limΔxexlna(eΔxlna−1)=Δx→0limΔxexlnaΔxlna=exlnalna=axlna
【例4.3.4】
y
=
x
α
,
α
∈
R
,
x
>
0
y=x^{\alpha},\alpha\in\mathbb{R},x>0
y=xα,α∈R,x>0
【解】
y
′
(
x
)
=
lim
Δ
x
→
0
(
x
+
Δ
x
)
α
−
x
α
Δ
x
=
lim
Δ
x
→
0
x
α
(
(
1
+
Δ
x
x
)
α
−
1
)
Δ
x
=
lim
Δ
x
→
0
x
α
⋅
α
Δ
x
x
Δ
x
=
α
x
α
−
1
y'(x)=\lim\limits_{\Delta x\to 0}\frac{(x+\Delta x)^{\alpha}-x^\alpha}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{x^\alpha ((1+\frac{\Delta x}{x})^\alpha - 1)}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{x^\alpha \cdot \alpha\frac{\Delta x}{x}}{\Delta x}=\alpha x^{\alpha - 1}
y′(x)=Δx→0limΔx(x+Δx)α−xα=Δx→0limΔxxα((1+xΔx)α−1)=Δx→0limΔxxα⋅αxΔx=αxα−1
【注】
y
=
x
n
,
n
=
1
,
2
,
3
,
.
.
.
,
x
∈
(
−
∞
,
+
∞
)
,
y
′
(
x
)
=
n
x
n
−
1
,
x
∈
(
−
∞
,
+
∞
)
y=x^n,n=1,2,3,...,x\in (-\infty,+\infty),y'(x)=nx^{n-1},x\in(-\infty,+\infty)
y=xn,n=1,2,3,...,x∈(−∞,+∞),y′(x)=nxn−1,x∈(−∞,+∞)
y
=
x
−
n
,
n
=
1
,
2
,
3
,
.
.
.
,
x
∈
(
−
∞
,
0
)
∪
(
0
,
+
∞
)
,
y
′
(
x
)
=
−
n
x
−
n
−
1
=
−
n
x
n
+
1
,
x
∈
(
−
∞
,
0
)
∪
(
0
,
+
∞
)
y=x^{-n},n=1,2,3,...,x\in(-\infty,0)\cup(0,+\infty),y'(x)=-nx^{-n-1}=\frac{-n}{x^{n+1}},x\in(-\infty,0)\cup(0,+\infty)
y=x−n,n=1,2,3,...,x∈(−∞,0)∪(0,+∞),y′(x)=−nx−n−1=xn+1−n,x∈(−∞,0)∪(0,+∞)
【例】
y
=
x
2
3
,
x
∈
(
−
∞
,
+
∞
)
,
y
′
(
x
)
=
2
3
x
−
1
3
=
2
3
x
3
,
x
∈
(
−
∞
,
0
)
∪
(
0
,
+
∞
)
y=x^{\frac{2}{3}},x\in(-\infty,+\infty),y'(x)=\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3\sqrt[3]{x}},x\in(-\infty,0)\cup(0,+\infty)
y=x32,x∈(−∞,+∞),y′(x)=32x−31=33x2,x∈(−∞,0)∪(0,+∞)，即它在
0
0
0点不可导。
【例】
y
=
x
1
2
,
x
∈
[
0
,
+
∞
)
,
y
′
(
x
)
=
1
2
x
,
x
∈
(
0
,
+
∞
)
y=x^{\frac{1}{2}},x\in[0,+\infty),y'(x)=\frac{1}{2\sqrt{x}},x\in(0,+\infty)
y=x21,x∈[0,+∞),y′(x)=2x1,x∈(0,+∞)，它在
0
0
0点右导数不存在。
4.3.1 导数四则运算
【定理4.3.1】
f
f
f和
g
g
g在同一区间可导，则
c
1
f
(
x
)
+
c
2
g
(
x
)
c_1f(x)+c_2g(x)
c1f(x)+c2g(x)也在该区间可导。且有
(
c
1
f
(
x
)
+
c
2
g
(
x
)
)
′
=
c
1
f
′
(
x
)
+
c
2
g
′
(
x
)
(c_1f(x)+c_2g(x))'=c_1f'(x)+c_2g'(x)
(c1f(x)+c2g(x))′=c1f′(x)+c2g′(x)
【证】
(
c
1
f
(
x
)
+
c
2
g
(
x
)
)
′
=
lim
Δ
x
→
0
(
c
1
f
(
x
+
Δ
x
)
+
c
2
g
(
x
+
Δ
x
)
)
−
(
c
1
f
(
x
)
+
c
2
g
(
x
)
)
Δ
x
=
lim
Δ
x
→
0
(
c
1
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
+
c
2
g
(
x
+
Δ
x
)
−
g
(
x
)
Δ
x
)
=
c
1
f
′
(
x
)
+
c
2
g
′
(
x
)
(c_1f(x)+c_2g(x))'=\lim\limits_{\Delta x\to 0}\frac{(c_1f(x+\Delta x)+c_2g(x+\Delta x))-(c_1f(x)+c_2g(x))}{\Delta x}=\lim\limits_{\Delta x\to 0}(c_1\frac{f(x+\Delta x)-f(x)}{\Delta x}+c_2\frac{g(x+\Delta x)-g(x)}{\Delta x})=c_1f'(x)+c_2g'(x)
(c1f(x)+c2g(x))′=Δx→0limΔx(c1f(x+Δx)+c2g(x+Δx))−(c1f(x)+c2g(x))=Δx→0lim(c1Δxf(x+Δx)−f(x)+c2Δxg(x+Δx)−g(x))=c1f′(x)+c2g′(x)
【例】
(
log
a
x
)
′
=
(
ln
x
ln
a
)
′
=
1
ln
a
⋅
1
x
=
1
x
ln
a
(\log_{a}x)'=(\frac{\ln x}{\ln a})'=\frac{1}{\ln a}\cdot\frac{1}{x}=\frac{1}{x\ln a}
(logax)′=(lnalnx)′=lna1⋅x1=xlna1
【定理4.3.2】
f
f
f和
g
g
g在同一区间可导，则
f
(
x
)
⋅
g
(
x
)
f(x)\cdot g(x)
f(x)⋅g(x)也在该区间可导，
(
f
(
x
)
g
(
x
)
)
′
=
f
′
(
x
)
g
(
x
)
+
f
(
x
)
g
′
(
x
)
(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)
(f(x)g(x))′=f′(x)g(x)+f(x)g′(x)
【证】
(
f
(
x
)
g
(
x
)
)
′
=
lim
Δ
x
→
0
f
(
x
+
Δ
x
)
g
(
x
+
Δ
x
)
−
f
(
x
)
g
(
x
)
Δ
x
=
lim
Δ
x
→
0
f
(
x
+
Δ
x
)
g
(
x
+
Δ
x
)
−
f
(
x
+
Δ
x
)
g
(
x
)
+
f
(
x
+
Δ
x
)
g
(
x
)
−
f
(
x
)
g
(
x
)
Δ
x
=
lim
Δ
x
→
0
f
(
x
+
Δ
x
)
(
g
(
x
+
Δ
x
)
−
g
(
x
)
)
Δ
x
+
lim
Δ
x
→
0
g
(
x
)
(
f
(
x
+
Δ
x
)
−
f
(
x
)
)
Δ
x
=
(
f
可导必连续
)
f
(
x
)
g
′
(
x
)
+
g
(
x
)
f
′
(
x
)
=
f
′
(
x
)
g
(
x
)
+
f
(
x
)
g
′
(
x
)
(f(x)g(x))'=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x)+f(x+\Delta x)g(x)-f(x)g(x)}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)(g(x+\Delta x)-g(x))}{\Delta x}+\lim\limits_{\Delta x\to 0}\frac{g(x)(f(x+\Delta x)-f(x))}{\Delta x}=(f可导必连续)f(x)g'(x)+g(x)f'(x)=f'(x)g(x)+f(x)g'(x)
(f(x)g(x))′=Δx→0limΔxf(x+Δx)g(x+Δx)−f(x)g(x)=Δx→0limΔxf(x+Δx)g(x+Δx)−f(x+Δx)g(x)+f(x+Δx)g(x)−f(x)g(x)=Δx→0limΔxf(x+Δx)(g(x+Δx)−g(x))+Δx→0limΔxg(x)(f(x+Δx)−f(x))=(f可导必连续)f(x)g′(x)+g(x)f′(x)=f′(x)g(x)+f(x)g′(x)
【例4.3.6】求
y
=
3
x
cos
x
y=3^{x}\cos x
y=3xcosx的导数
【解】
y
′
(
x
)
=
3
x
(
ln
3
)
⋅
cos
x
−
3
x
sin
x
=
3
x
(
(
ln
3
)
⋅
cos
x
−
sin
x
)
y'(x)=3^x(\ln 3)\cdot\cos x-3^x\sin x=3^x((\ln 3)\cdot \cos x - \sin x)
y′(x)=3x(ln3)⋅cosx−3xsinx=3x((ln3)⋅cosx−sinx)
【例4.3.7】求
y
=
sin
x
x
y=\frac{\sin x}{x}
y=xsinx的导数。
【解】
y
′
(
x
)
=
1
x
cos
x
+
(
−
1
x
2
)
sin
x
=
x
cos
x
−
sin
x
x
2
y'(x)=\frac{1}{x}\cos x+(-\frac{1}{x^2})\sin x=\frac{x\cos x - \sin x}{x^2}
y′(x)=x1cosx+(−x21)sinx=x2xcosx−sinx
【定理4.3.3】设
g
(
x
)
g(x)
g(x)在某一个区间可导，
g
(
x
)
≠
0
g(x)\ne 0
g(x)=0，则
1
g
(
x
)
\frac{1}{g(x)}
g(x)1也在该区间可导，并且
(
1
g
(
x
)
)
′
=
−
g
′
(
x
)
g
2
(
x
)
(\frac{1}{g(x)})'=\frac{-g'(x)}{g^{2}(x)}
(g(x)1)′=g2(x)−g′(x)
【证】
(
1
g
(
x
)
)
′
=
lim
Δ
x
→
0
1
g
(
x
+
Δ
x
)
−
1
g
(
x
)
Δ
x
=
lim
Δ
x
→
0
−
(
g
(
x
+
Δ
x
)
−
g
(
x
)
)
Δ
x
g
(
x
)
g
(
x
+
Δ
x
)
=
(
g
可导必连续
)
−
g
′
(
x
)
g
2
(
x
)
(\frac{1}{g(x)})'=\lim\limits_{\Delta x\to 0}\frac{\frac{1}{g(x+\Delta x)}-\frac{1}{g(x)}}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{-(g(x+\Delta x)-g(x))}{\Delta xg(x)g(x+\Delta x)}=(g可导必连续)\frac{-g'(x)}{g^{2}(x)}
(g(x)1)′=Δx→0limΔxg(x+Δx)1−g(x)1=Δx→0limΔxg(x)g(x+Δx)−(g(x+Δx)−g(x))=(g可导必连续)g2(x)−g′(x)
【注】
d
f
=
f
′
(
x
)
d
x
⇔
d
y
d
x
=
f
′
(
x
)
,
d
(
1
g
(
x
)
)
=
−
g
′
(
x
)
g
2
(
x
)
d
x
=
−
d
g
(
x
)
g
2
(
x
)
df=f'(x)dx\Leftrightarrow \frac{dy}{dx}=f'(x),d(\frac{1}{g(x)})=\frac{-g'(x)}{g^{2}(x)}dx=-\frac{dg(x)}{g^{2}(x)}
df=f′(x)dx⇔dxdy=f′(x),d(g(x)1)=g2(x)−g′(x)dx=−g2(x)dg(x)
【例】
(
sin
x
)
′
=
cos
x
,
(
cos
x
)
′
=
−
sin
x
(\sin x)'=\cos x,(\cos x)'=-\sin x
(sinx)′=cosx,(cosx)′=−sinx，求
(
sec
x
)
′
(\sec x)'
(secx)′
【解】
(
sec
x
)
′
=
(
1
cos
x
)
′
=
−
(
−
sin
x
)
cos
2
x
=
sin
x
cos
2
x
=
tan
x
sec
x
(\sec x)'=(\frac{1}{\cos x})'=-\frac{(-\sin x)}{\cos^2 x}=\frac{\sin x}{\cos ^ 2 x}=\tan x\sec x
(secx)′=(cosx1)′=−cos2x(−sinx)=cos2xsinx=tanxsecx
同理
(
csc
x
)
′
=
−
cot
x
csc
x
(\csc x)'=-\cot x\csc x
(cscx)′=−cotxcscx
【推论】
f
,
g
f,g
f,g在同一区间可导，
g
(
x
)
≠
0
g(x)\ne 0
g(x)=0，则
f
(
x
)
g
(
x
)
\frac{f(x)}{g(x)}
g(x)f(x)在该区间可导，并且
(
f
(
x
)
g
(
x
)
)
′
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
g
2
(
x
)
(\frac{f(x)}{g(x)})'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}
(g(x)f(x))′=g2(x)f′(x)g(x)−f(x)g′(x)
【证】可以由前面的两个公式推出
(
f
(
x
)
g
(
x
)
)
′
=
(
f
(
x
)
⋅
1
g
(
x
)
)
′
=
f
′
(
x
)
1
g
(
x
)
+
f
(
x
)
⋅
(
−
g
′
(
x
)
g
2
(
x
)
)
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
g
2
(
x
)
(\frac{f(x)}{g(x)})'=(f(x)\cdot\frac{1}{g(x)})'=f'(x)\frac{1}{g(x)}+f(x)\cdot(-\frac{g'(x)}{g^2(x)})=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}
(g(x)f(x))′=(f(x)⋅g(x)1)′=f′(x)g(x)1+f(x)⋅(−g2(x)g′(x))=g2(x)f′(x)g(x)−f(x)g′(x)
【例】
(
tan
x
)
′
=
(
sin
x
cos
x
)
′
=
cos
2
x
−
(
sin
x
⋅
(
−
sin
x
)
)
cos
2
x
=
cos
2
x
+
sin
2
x
cos
2
x
=
1
cos
2
x
=
sec
2
x
(\tan x)'=(\frac{\sin x}{\cos x})'=\frac{\cos^2 x-(\sin x\cdot (-\sin x))}{\cos ^2 x}=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}=\frac{1}{\cos^2 x}=\sec^2 x
(tanx)′=(cosxsinx)′=cos2xcos2x−(sinx⋅(−sinx))=cos2xcos2x+sin2x=cos2x1=sec2x
同理
(
cot
x
)
′
=
−
csc
2
x
(\cot x)'=-\csc^ 2 x
(cotx)′=−csc2x
4.3.2 反函数求导法则
【定理4.3.4】【反函数求导定理】
f
(
x
)
f(x)
f(x)在
(
a
,
b
)
(a,b)
(a,b)连续且严格单调并且可导，
f
′
(
x
)
≠
0
f'(x)\ne 0
f′(x)=0，
α
=
min
{
f
(
a
+
)
,
f
(
b
−
)
}
,
β
=
{
f
(
a
+
)
,
f
(
b
−
)
}
\alpha =\min\{f(a+),f(b-)\},\beta=\{f(a+),f(b-)\}
α=min{f(a+),f(b−)},β={f(a+),f(b−)}，则
f
−
1
(
y
)
f^{-1}(y)
f−1(y)在
(
α
,
β
)
(\alpha,\beta)
(α,β)上可导且
(
f
−
1
(
y
)
)
′
=
1
f
′
(
x
)
,
(
x
=
f
−
1
(
y
)
)
(f^{-1}(y))'=\frac{1}{f'(x)},(x=f^{-1}(y))
(f−1(y))′=f′(x)1,(x=f−1(y))
【证】
y
=
f
(
x
)
,
y
+
Δ
y
=
f
(
x
+
Δ
x
)
,
x
=
f
−
1
(
y
)
,
f
−
1
(
y
+
Δ
y
)
=
x
+
Δ
x
y=f(x),y+\Delta y=f(x+\Delta x),x=f^{-1}(y),f^{-1}(y+\Delta y)=x+\Delta x
y=f(x),y+Δy=f(x+Δx),x=f−1(y),f−1(y+Δy)=x+Δx
由于
f
f
f是严格单调的，
Δ
x
≠
0
⇔
Δ
y
≠
0
\Delta x\ne 0\Leftrightarrow \Delta y \ne 0
Δx=0⇔Δy=0，由
f
f
f的连续性，
Δ
x
→
0
⇔
Δ
y
→
0
\Delta x\to 0\Leftrightarrow\Delta y\to 0
Δx→0⇔Δy→0
实际上
(
f
−
1
(
y
)
)
′
=
lim
Δ
y
→
0
f
−
1
(
y
+
Δ
y
)
−
f
−
1
(
y
)
Δ
y
=
lim
Δ
y
→
0
x
+
Δ
x
−
x
Δ
y
=
lim
Δ
y
→
0
Δ
x
f
(
x
+
Δ
x
)
−
y
=
lim
Δ
x
→
0
Δ
x
f
(
x
+
Δ
x
)
−
f
(
x
)
=
lim
Δ
x
→
0
1
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
=
1
f
′
(
x
)
(f^{-1}(y))'=\lim\limits_{\Delta y\to 0}\frac{f^{-1}(y+\Delta y)-f^{-1}(y)}{\Delta y}=\lim\limits_{\Delta y\to 0}\frac{x + \Delta x - x}{\Delta y}=\lim\limits_{\Delta y\to 0}\frac{\Delta x}{f(x+\Delta x)-y}=\lim\limits_{\Delta x\to 0}\frac{\Delta x}{f(x+\Delta x)-f(x)}=\lim\limits_{\Delta x\to 0}\frac{1}{\frac{f(x+\Delta x)-f(x)}{\Delta x}}=\frac{1}{f'(x)}
(f−1(y))′=Δy→0limΔyf−1(y+Δy)−f−1(y)=Δy→0limΔyx+Δx−x=Δy→0limf(x+Δx)−yΔx=Δx→0limf(x+Δx)−f(x)Δx=Δx→0limΔxf(x+Δx)−f(x)1=f′(x)1（因为
Δ
x
\Delta x
Δx趋于0但是不等于0，可以除下来）
【例】
(
arctan
x
)
′
=
1
(
tan
y
)
′
=
1
sec
2
y
=
1
1
+
tan
2
y
=
1
1
+
x
2
,
(
y
=
arctan
x
,
tan
y
=
x
)
(\arctan x)'=\frac{1}{(\tan y)'}=\frac{1}{\sec^2 y}=\frac{1}{1+\tan^2 y}=\frac{1}{1+x^2},(y=\arctan x,\tan y =x)
(arctanx)′=(tany)′1=sec2y1=1+tan2y1=1+x21,(y=arctanx,tany=x)
同理
(
arccot
x
)
′
=
−
1
1
+
x
2
(\text{arccot} x)'=-\frac{1}{1+x^2}
(arccotx)′=−1+x21
【例】
(
arcsin
x
)
′
=
1
(
sin
y
)
′
=
1
cos
y
=
1
1
−
sin
2
y
=
1
1
−
x
2
,
(
x
=
sin
y
)
(\arcsin x)'=\frac{1}{(\sin y)'}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-\sin^2 y}}=\frac{1}{\sqrt{1-x^2}},(x=\sin y)
(arcsinx)′=(siny)′1=cosy1=1−sin2y1=1−x21,(x=siny)
同理
(
arccos
x
)
′
=
−
1
1
−
x
2
(\arccos x)'=-\frac{1}{\sqrt{1-x^2}}
(arccosx)′=−1−x21
原文地址：https://blog.csdn.net/qq_30204431/article/details/142678053
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