本文探讨了极限运算法则、Stolz定理、夹逼定理等在无穷级数中的应用，通过具体例子解析了1/n和n/n+1等序列的极限，以及涉及对数、三角函数和指数函数的极限计算。还涉及了导数及其求解技巧，如arcsin、arccos和arctan的导数计算。
关于极限
二项式定理
(a+b)n=Cn0an+Cn1an−1b+Cn2an−2b2+⋯+Cnn−1abn−1+Cnnbn=∑i=0nCnian−ibi (a + b)^n = C_n^0a^n + C_n^1a^{n-1}b + C_n^2a^{n-2}b^2 + \cdots + C_n^{n-1}ab^{n-1} + C_n^nb^n = \sum_{i=0}^{n}C_n^i a^{n-i}b^i (a+b)n=Cn0an+Cn1an−1b+Cn2an−2b2+⋯+Cnn−1abn−1+Cnnbn=i=0∑nCnian−ibi
极限运算法则(们)
limn→∞(xn+yn)=limn→∞xn+limn→∞ynlimn→∞(xnyn)=limn→∞xn⋅limn→∞ynlimn→∞(xnyn)=limn→∞xnlimn→∞ynlimn→∞xnyn=(limn→∞xn)limn→∞yn \begin{aligned} & \lim_{n\to \infty}(x_n+y_n) = \lim_{n\to \infty}x_n + \lim_{n\to \infty} y_n \\ \\ & \lim_{n\to \infty}(x_ny_n) = \lim_{n\to \infty}x_n \cdot \lim_{n\to \infty}y_n\\ \\ & \lim_{n\to \infty}\left( \frac{x_n}{y_n} \right) = \frac{\lim_{n\to \infty}x_n}{\lim_{n\to \infty}y_n} \\ \\ &\lim_{n\to \infty}x_n^{y_n} = \left(\lim_{n\to \infty}x_n\right)^{\lim_{n\to \infty}y_n} \end{aligned} n→∞lim(xn+yn)=n→∞limxn+n→∞limynn→∞lim(xnyn)=n→∞limxn⋅n→∞limynn→∞lim(ynxn)=limn→∞ynlimn→∞xnn→∞limxnyn=(n→∞limxn)limn→∞yn
Stolz 定理
两个数列 {xn}\lbrace x_n \rbrace{xn} 和 {yn}\lbrace y_n \rbrace{yn}，其中 yny_nyn 是无穷大量。如果：
limn→∞xn−xn−1yn−yn−1=a \lim_{n\to \infty}\frac{x_n - x_{n-1}}{y_n-y_{n-1}} = a n→∞limyn−yn−1xn−xn−1=a
则：
limn→∞xnyn=a \lim_{n\to \infty} \frac{x_n}{y_n} = a n→∞limynxn=a
夹逼定理
三个数列 {xn}\lbrace x_n \rbrace{xn} , {yn}\lbrace y_n \rbrace{yn} 和 {zn}\lbrace z_n \rbrace{zn}。若：
xn≤yn≤zn且limn→∞xn=limn→∞zn=a x_n \leq y_n \leq z_n \qquad 且 \qquad \lim_{n\to \infty}x_n = \lim_{n\to \infty}z_n = a xn≤yn≤zn且n→∞limxn=n→∞limzn=a
则：
limn→∞yn=a \lim_{n\to \infty}y_n = a n→∞limyn=a
题(们)
（1）
limn→∞(1+12+13+⋯+1n)1n≤limn→∞n1n=1\lim_{n\to \infty}(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n})^{\frac{1}{n}} \leq \lim_{n\to \infty}n^{\frac{1}{n}} = 1 n→∞lim(1+21+31+⋯+n1)n1≤n→∞limnn1=1
limn→∞(1+12+13+⋯+1n)1n≥limn→∞11n=1 \lim_{n\to \infty}(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n})^{\frac{1}{n}} \geq \lim_{n\to \infty}1^{\frac{1}{n}} = 1 n→∞lim(1+21+31+⋯+n1)n1≥n→∞lim1n1=1
∴limn→∞(1+12+13+⋯+1n)1n=1 \therefore \lim_{n\to \infty}(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n})^{\frac{1}{n}} = 1 ∴n→∞lim(1+21+31+⋯+n1)n1=1
（2）
limn→∞(1n+1+1n+2+⋯+1n+n)≤limn→∞nn+n=limn→∞11+1n=1 \lim_{n\to \infty}(\frac{1}{n + \sqrt{1}} + \frac{1}{n + \sqrt{2}} + \cdots +\frac{1}{n + \sqrt{n}}) \leq \lim_{n\to \infty} \frac{n}{n + \sqrt{n}} = \lim_{n\to \infty}\frac{1}{1+\frac{1}{\sqrt{n}}} = 1 n→∞lim(n+11+n+21+⋯+n+n1)≤n→∞limn+nn=n→∞lim1+n11=1
limn→∞(nn+1+nn+2+⋯+nn+n)≥limn→∞nn+1=11+1n=1 \lim_{n\to \infty}({\frac{n}{n +\sqrt{1}} + \frac{n}{n +\sqrt{2}} +\cdots+ \frac{n}{n +\sqrt{n}}}) \geq \lim_{n\to \infty}\frac{n}{n+1} = \frac{1}{1+\frac{1}{n}} = 1 n→∞lim(n+1n+n+2n+⋯+n+nn)≥n→∞limn+1n=1+n11=1
∴limn→∞(1n+1+1n+2+⋯+1n+n)=1 \therefore \lim_{n\to \infty}(\frac{1}{n+\sqrt{1}} + \frac{1}{n +\sqrt{2}} + \cdots + \frac{1}{n +\sqrt{n}}) = 1 ∴n→∞lim(n+11+n+21+⋯+n+n1)=1
（3）
limn→∞∑k=n2(n+1)21k≤limn→∞(n+1)2−n2+1n+1=limn→∞2=2 \lim_{n\to \infty}\sum_{k = n^2}^{(n+1)^2}\frac{1}{\sqrt{k}} \leq \lim_{n\to \infty}\frac{(n+1)^2-n^2+1}{n+1} = \lim_{n\to \infty}2 = 2 n→∞limk=n2∑(n+1)2k1≤n→∞limn+1(n+1)2−n2+1=n→∞lim2=2
limn→∞∑k=n2(n+1)21k≥(n+1)2−n2+1n=2limn→∞nn+1=2limn→∞11+1n=2 \lim_{n\to \infty}\sum_{k = n^2}^{(n+1)^2}\frac{1}{\sqrt{k}} \geq \frac{(n+1)^2-n^2+1}{n} = 2\lim_{n\to \infty}\frac{n}{n+1} = 2\lim_{n\to \infty}\frac{1}{1+\frac{1}{n}} = 2 n→∞limk=n2∑(n+1)2k1≥n(n+1)2−n2+1=2n→∞limn+1n=2n→∞lim1+n11=2
（4）
limn→∞1⋅3⋅5⋯(2n−1)2⋅4⋅6⋯(2n)≥limn→∞(12)n=0 \lim_{n\to \infty}\frac{1\cdot 3\cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \geq \lim_{n\to \infty}(\frac{1}{2})^n = 0 n→∞lim2⋅4⋅6⋯(2n)1⋅3⋅5⋯(2n−1)≥n→∞lim(21)n=0
令G=∏i=1n2i−12i，P=∏i=1n2i2i+1 令 G = \prod_{i=1}^{n}\frac{2i-1}{2i}，P = \prod_{i=1}^{n}\frac{2i}{2i+1} 令G=i=1∏n2i2i−1，P=i=1∏n2i+12i
G<P且GP=12n+1∴G<12n+1→limn→∞G≤limn→∞12n+1=0 G < P 且 GP = \frac{1}{2n+1} \therefore G < \frac{1}{\sqrt{2n+1}} \rightarrow \lim_{n\to \infty}G \leq \lim_{n\to \infty}\frac{1}{\sqrt{2n+1}} = 0 G<P且GP=2n+11∴G<2n+11→n→∞limG≤n→∞lim2n+11=0
limn→∞1⋅3⋅5⋯(2n−1)2⋅4⋅6⋯(2n)=0 \lim_{n\to \infty}\frac{1\cdot 3\cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} = 0 n→∞lim2⋅4⋅6⋯(2n)1⋅3⋅5⋯(2n−1)=0
（1）
limn→∞3n2+4n−1n2+1=3+limn→∞n−1n2+1=3+limn→∞1−1nn+1n=3 \lim_{n\to \infty}\frac{3n^2+4n-1}{n^2+1} = 3 + \lim_{n\to \infty}\frac{n-1}{n^2+1} = 3 + \lim_{n\to \infty}\frac{1-\frac{1}{n}}{n+\frac{1}{n}} = 3 n→∞limn2+13n2+4n−1=3+n→∞limn2+1n−1=3+n→∞limn+n11−n1=3
（2）
limn→∞n3+2n2−3n+12n3−n+3=limn→∞1+2n+3n2+1n32−1n+1n3=12 \lim_{n\to \infty}\frac{n^3+2n^2-3n+1}{2n^3-n+3} = \lim_{n\to \infty}\frac{1 + \frac{2}{n} + \frac{3}{n^2} + \frac{1}{n^3}}{2 - \frac{1}{n} + \frac{1}{n^3}} = \frac{1}{2} n→∞lim2n3−n+3n3+2n2−3n+1=n→∞lim2−n1+n311+n2+n23+n31=21
（3）
limn→∞3n+n33n+1+(n+1)3=limn→∞1+n33n3+(n+1)33n=13（limn→∞nkan=0证明见8） \lim_{n\to \infty}\frac{3^n + n^3}{3^{n+1}+(n+1)^3} = \lim_{n\to \infty}\frac{1+\frac{n^3}{3^n}}{3 + \frac{(n+1)^3}{3^n}} = \frac{1}{3} （\lim_{n\to \infty}\frac{n^k}{a^n} = 0 证明见 8）n→∞lim3n+1+(n+1)33n+n3=n→∞lim3+3n(n+1)31+3nn3=31（n→∞limannk=0证明见8）
（4）
limn→∞(n2+1n−1)sinnπ2=[(limn→∞n2+1n)−1]limn→∞sinnπ2=0 \lim_{n\to \infty}(\sqrt[n]{n^2+1}-1)\sin\frac{n\pi}{2} = \Bigg[\Big(\lim_{n\to \infty}\sqrt[n]{n^2+1}\Big)-1\Bigg]\lim_{n\to \infty}\sin\frac{n\pi}{2} = 0 n→∞lim(nn2+1−1)sin2nπ=[(n→∞limnn2+1)−1]n→∞limsin2nπ=0
（5）
limn→∞n(n+1−n)=limn→∞(n2+n−n)(n2+n+n)n2+n+n=limn→∞nn2+n+n=limn→∞11+1+1n=12
\begin{aligned}
&\lim_{n\to \infty}\sqrt{n}(\sqrt{n+1} - \sqrt{n}) = \lim_{n\to \infty}\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n} + n)}{\sqrt{n^2+n}+n}\\
= &\lim_{n\to \infty}\frac{n}{\sqrt{n^2+n}+n} = \lim_{n\to \infty}\frac{1}{1 + \sqrt{1+\frac{1}{n}}} = \frac{1}{2}
\end{aligned}
=n→∞limn(n+1−n)=n→∞limn2+n+n(n2+n−n)(n2+n+n)n→∞limn2+n+nn=n→∞lim1+1+n11=21
（6）
limn→∞(1−122)(1−132)⋯(1−1n2)=limn→∞22−122⋅32−132⋅42−142⋯n2−1n2=limn→∞(2+1)(2−1)(3+1)(3−1)(4+1)(4−1)⋯(n+1)(n−1)22⋅32⋅42⋯n2=limn→∞n+12n=limn→∞1+1n2=12
\begin{aligned}
&\lim_{n\to \infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})\cdots (1-\frac{1}{n^2})\\
= & \lim_{n\to \infty}\frac{2^2-1}{2^2} \cdot \frac{3^2-1}{3^2} \cdot \frac{4^2-1}{4^2} \cdots \frac{n^2-1}{n^2}\\
= &\lim_{n\to \infty}\frac{(2+1)(2-1)(3+1)(3-1)(4+1)(4-1) \cdots (n+1)(n-1)}{2^2\cdot3^2\cdot4^2\cdots n^2} \\
= &\lim_{n\to \infty}\frac{n+1}{2n} = \lim_{n\to \infty}\frac{1+\frac{1}{n}}{2} = \frac{1}{2}
\end{aligned}
===n→∞lim(1−221)(1−321)⋯(1−n21)n→∞lim2222−1⋅3232−1⋅4242−1⋯n2n2−1n→∞lim22⋅32⋅42⋯n2(2+1)(2−1)(3+1)(3−1)(4+1)(4−1)⋯(n+1)(n−1)n→∞lim2nn+1=n→∞lim21+n1=21
（7）
limn→∞n(n2+14−n+1)=limn→∞nn2+14+n2+2n+14=limn→∞nn2+1−n2+2n+1n2+14+n2+2n+14=limn→∞n−2n(n2+14+n2+2n+14)(n2+1+n2+2n+1)=limn→∞−2n32(n2+14+n2+2n+14)(n2+1+n2+2n+1)=limn→∞−2n324n32=−12
\begin{aligned}
& \lim_{n\to \infty}\sqrt{n}(\sqrt[4]{n^2+1}- \sqrt{n+1})\\
= & \lim_{n\to \infty}\sqrt{n}{\sqrt[4]{n^2+1} + \sqrt[4]{n^2+2n+1}}\\
= & \lim_{n\to \infty}\sqrt{n}\frac{\sqrt{n^2+1} - \sqrt{n^2+2n+1}}{\sqrt[4]{n^2+1}+\sqrt[4]{n^2+2n+1}}\\
= & \lim_{n\to \infty}\sqrt{n}\frac{-2n}{\left( \sqrt[4]{n^2+1}+\sqrt[4]{n^2+2n+1} \right) \left( \sqrt{n^2+1}+\sqrt{n^2+2n+1} \right) }\\
= & \lim_{n\to \infty} \frac{-2n^{\frac{3}{2}}}{\left( \sqrt[4]{n^2+1}+\sqrt[4]{n^2+2n+1} \right) \left( \sqrt{n^2+1}+\sqrt{n^2+2n+1} \right) }\\
= & \lim_{n\to \infty}\frac{-2n^{\frac{3}{2}}}{4n^{\frac32}} = -\frac 12
\end{aligned}
=====n→∞limn(4n2+1−n+1)n→∞limn4n2+1+4n2+2n+1n→∞limn4n2+1+4n2+2n+1n2+1−n2+2n+1n→∞limn(4n2+1+4n2+2n+1)(n2+1+n2+2n+1)−2nn→∞lim(4n2+1+4n2+2n+1)(n2+1+n2+2n+1)−2n23n→∞lim4n23−2n23=−21
（8）
limn→∞(12+322+⋯+2n−12n) \lim_{n\to \infty}(\frac{1}{2} + \frac{3}{2^2} + \cdots + \frac{2n-1}{2^n}) n→∞lim(21+223+⋯+2n2n−1)
- 证明：若 limn→∞an=a\lim_{n\to \infty} a_n = alimn→∞an=a 则 limn→∞a1+a2+a3+⋯+ann=a\lim_{n\to \infty}\frac{a_1+a_2+a_3+\cdots + a_n}{n} = alimn→∞na1+a2+a3+⋯+an=a
令 xn=a1+⋯+an，yn=nx_n = a_1+\cdots+a_n，y_n = nxn=a1+⋯+an，yn=n
limn→∞11+⋯+ann=limn→∞xn+1−xnyn+1−yn=limn→∞an+11=a \begin{aligned} \lim_{n\to \infty}\frac{1_1+\cdots+a_n}{n} = \lim_{n\to \infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n} = \lim_{n\to \infty}\frac{a_{n+1}}{1} = a \end{aligned} n→∞limn11+⋯+an=n→∞limyn+1−ynxn+1−xn=n→∞lim1an+1=a
证毕
- 设 an>0a_n > 0an>0， 且 limn→∞an=a\lim_{n\to \infty}a_n = alimn→∞an=a，证明：limn→∞a1a2⋯ann=a\lim_{n\to \infty}\sqrt[n]{a_1a_2\cdots a_n} = alimn→∞na1a2⋯an=a
limn→∞a1⋯ann≤limn→∞a1+⋯+ann=a \lim_{n\to \infty}\sqrt[n]{a_1\cdots a_n} \leq \lim_{n\to \infty}\frac{a_1 + \cdots + a_n}{n} = a n→∞limna1⋯an≤n→∞limna1+⋯+an=a
limn→∞aa⋯ann≥limn→∞an=a \lim_{n\to \infty}\sqrt[n]{a_a\cdots a_n} \geq \lim_{n\to \infty}a_n = a n→∞limnaa⋯an≥n→∞liman=a
∴limn→∞a1a2⋯ann=a \therefore \lim_{n\to \infty}\sqrt[n]{a_1a_2\cdots a_n} = a ∴n→∞limna1a2⋯an=a
证毕
- 若 an>0a_n > 0an>0 （n = 1， 2， 3…），且 limn→∞an+1an=a\lim_{n\to \infty}\frac{a_{n+1}}{a_n} = alimn→∞anan+1=a，求证 limn→∞ann=a\lim_{n\to \infty}\sqrt[n]{a_n} = alimn→∞nan=a
- 求 limn→∞12+33+52+⋯+(2n+1)2n3\lim_{n\to \infty}\frac{1^2 + 3^3 + 5^2 + \cdots + (2n+1)^2}{n^3}limn→∞n312+33+52+⋯+(2n+1)2
- 用 Stolz 定理求 limn→∞logann\lim_{n\to \infty}\frac{\log_an}{n}limn→∞nlogan 和 limn→∞nkan\lim_{n\to \infty}\frac{n^k}{a^n}limn→∞annk
limn→∞logann=limn→∞loga(n+1)−logann+1−n=limn→∞loga(n+1n)=limn→∞loga(1+1n)=0 \begin{aligned} \lim_{n\to \infty}\frac{\log_an}{n} = &\lim_{n\to \infty}\frac{\log_a(n+1)-\log_an}{n+1-n} \\ = & \lim_{n\to \infty}\log_a(\frac{n+1}{n})\\ = & \lim_{n\to \infty}\log_a(1+\frac{1}{n}) = 0 \end{aligned} n→∞limnlogan===n→∞limn+1−nloga(n+1)−logann→∞limloga(nn+1)n→∞limloga(1+n1)=0
limn→∞nkan=limn→∞nk−(n−1)kan−an−1=limn→∞Ck1nk−1−Ck2nk−2+Ck3nk−3−⋯+1an−an−1 \begin{aligned} &\lim_{n\to \infty}\frac{n^k}{a^n} = \lim_{n\to \infty}\frac{n^k-(n-1)^k}{a^{n}-a^{n-1}}\\ = & \lim_{n\to \infty}\frac{C_k^1n^{k-1}-C_k^2n^{k-2}+C_k^3n^{k-3}-\cdots+1}{a^n-a^{n-1}} \end{aligned} =n→∞limannk=n→∞liman−an−1nk−(n−1)kn→∞liman−an−1Ck1nk−1−Ck2nk−2+Ck3nk−3−⋯+1
- 设 {xn}\lbrace x_n \rbrace{xn} 是无穷大量， ∣yn∣≥δ>0|y_n| \geq \delta > 0∣yn∣≥δ>0，证明 {xnyn}\lbrace x_ny_n \rbrace{xnyn} 是无穷大量。
导数
常见的导数(们)
(C)′=0(sinx)′=cosx(cosx)′=−sinx(lnx)′=1x(logax)′=1xlna(ex)′=ex(ax)′=axlna(xn)′=nxn−1 \begin{aligned} &(C)' = 0\\ &(\sin x)' = cosx\\ &(\cos x)' = -sinx\\ &(\ln x)' = \frac 1x\\ &(\log_ax)' = \frac{1}{xlna}\\ &(e^x)' = e^x\\ &(a^x)' = a^xlna\\ &(x^n)' = nx^{n-1} \end{aligned} (C)′=0(sinx)′=cosx(cosx)′=−sinx(lnx)′=x1(logax)′=xlna1(ex)′=ex(ax)′=axlna(xn)′=nxn−1
求导法则(们)
d(f(x)+g(x))dx=df(x)dx+dg(x)dxd(f(x)g(x))dx=df(x)dxg(x)+dg(x)dxf(x)d(f(x)g(x))dx=df(x)dxg(x)−dg(x)dxf(x)g2(x)df(g(x))dx=df(g(x))dg(x)⋅dg(x)dx
\begin{aligned}
&\frac{d(f(x)+g(x))}{dx} = \frac{df(x)}{dx} + \frac{dg(x)}{dx}\\ \\
&\frac{d\left(f(x)g(x)\right)}{dx} = \frac{df(x)}{dx}g(x) + \frac{dg(x)}{dx}f(x)\\ \\
&\frac{d\left(\frac{f(x)}{g(x)}\right)}{dx} = \frac{\frac{df(x)}{dx}g(x) - \frac{dg(x)}{dx}f(x)}{g^2(x)} \\ \\
&\frac{df(g(x))}{dx} = \frac{df(g(x))}{dg(x)} \cdot \frac{dg(x)}{dx}
\end{aligned}
dxd(f(x)+g(x))=dxdf(x)+dxdg(x)dxd(f(x)g(x))=dxdf(x)g(x)+dxdg(x)f(x)dxd(g(x)f(x))=g2(x)dxdf(x)g(x)−dxdg(x)f(x)dxdf(g(x))=dg(x)df(g(x))⋅dxdg(x)
需要注意的是，复合函数求导法则不适用于指数形式，比如：xxx^xxx 的导数应该这样求：
(xx)′=(eln(xx))′=(exlnx)′=exlnx⋅(lnx+1)
\begin{aligned}
(x^x)' = \left(e^{\ln(x^x)}\right)' = \left( e^{x\ln x} \right)' = e^{x\ln x} \cdot (\ln x + 1)
\end{aligned}
(xx)′=(eln(xx))′=(exlnx)′=exlnx⋅(lnx+1)
题(们)
当 a 为何值时，直线 y=xy = xy=x 能与 y=logaxy = \log_axy=logax 相切，切点在哪里。
因为相切所以切点切线斜率为 1，即：
(logax)′=1xlna=1 \left( \log_ax \right)' = \frac{1}{x\ln a} = 1 (logax)′=xlna1=1
又因为切点在 y=xy = xy=x 上，所以横纵坐标相等，即：
x=logax x = \log_ax x=logax
联立得：
{1x0lna=1x0=logax0∵1x0lna=1∴x0=1lna∵x0=logax0∴x0=lnx0lna∴lnx0=1→x0=e→lna=1e→a=e1e
\begin{aligned}
&\begin{cases}
\frac{1}{x_0\ln a} = 1 \\
x_0 = \log_a x_0
\end{cases}\\
& \because \frac{1}{x_0\ln a} = 1 \quad \therefore x_0 = \frac{1}{\ln a}\\
& \because x_0 = \log_ax_0 \quad \therefore x_0 = \frac{\ln x_0}{\ln a}\\
&\therefore \ln x_0 = 1 \rightarrow x_0 = e \rightarrow \ln a = \frac{1}{e} \rightarrow a = e^{\frac{1}{e}}
\end{aligned}
{x0lna1=1x0=logax0∵x0lna1=1∴x0=lna1∵x0=logax0∴x0=lnalnx0∴lnx0=1→x0=e→lna=e1→a=ee1
所以当 a=e1ea = e^{\frac{1}{e}}a=ee1 时， y=logaxy = \log_axy=logax 与 y=xy=xy=x 相切，切点为 (e,e)(e, e)(e,e)。
求 arcsinx\arcsin xarcsinx 的导数：
我们知道如果 y=arcsinxy = \arcsin xy=arcsinx 则 x=sinyx = \sin yx=siny。等式两边同时求导，得到：
y′cosy=1→y′=1cosy=11−sin2y y' \cos y = 1 \rightarrow y' = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^2 y}} y′cosy=1→y′=cosy1=1−sin2y1
又因为 y=arcsinxy = \arcsin xy=arcsinx，所以：
y′=11−sin2(arcsinx)=11−x2 y' = \frac{1}{\sqrt{1 - \sin^2(\arcsin x)}} = \frac{1}{\sqrt{1-x^2}} y′=1−sin2(arcsinx)1=1−x21
求 arccosx\arccos xarccosx的导数：
因为 y=arccosxy = \arccos xy=arccosx，我们就有 x=cosyx = \cos yx=cosy。两边同时求导得到：
1=−y′siny→y′=−1siny=−11−cos2y 1 = -y' \sin y \rightarrow y' = -\frac{1}{\sin y} = -\frac{1}{\sqrt{1-\cos^2 y}} 1=−y′siny→y′=−siny1=−1−cos2y1
又因为 y=arccosxy = \arccos xy=arccosx 所以：
y′=−11−cos2(arccosx)=−11−x2 y' = -\frac{1}{\sqrt{1-\cos^2(\arccos x)}} = -\frac{1}{\sqrt{1-x^2}} y′=−1−cos2(arccosx)1=−1−x21
求 arctanx\arctan xarctanx 的导数：
设 y=arctanxy = \arctan xy=arctanx 则 x=tanyx = \tan yx=tany两边同时求导得到：
1=y′(tany)′=y′(sinycosy)′=y′(cos2y+sin2ycos2y)=y′cos2y 1 = y' (\tan y)' = y' \left(\frac{\sin y}{\cos y}\right)' = y' \left( \frac{\cos^2 y + \sin^2 y}{\cos^2 y} \right) = \frac{y'}{\cos^2 y} 1=y′(tany)′=y′(cosysiny)′=y′(cos2ycos2y+sin2y)=cos2yy′
所以：
y′=cos2y=cos2ysin2y+cos2y=11+tan2y y' = \cos^2 y = \frac{\cos^2 y}{\sin^2 y + \cos^2 y} = \frac{1}{1 + \tan^2 y} y′=cos2y=sin2y+cos2ycos2y=1+tan2y1
又因为 y=arctanxy = \arctan xy=arctanx 所以：
y′=11+tan2(arctanx)=11+x2 y' = \frac{1}{1 +\tan^2(\arctan x)} = \frac{1}{1 + x^2} y′=1+tan2(arctanx)1=1+x21
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