前段时间复习完了高数第二章的内容，我参考《复习全书·基础篇》和老师讲课的内容对这一章的知识点进行了整理，形成了这篇笔记，方便在移动设备上进行访问和后续的补充修改。

连续与可导

连续与可微

可导与可微(在一元函数中)

注：在多元函数中，可导（偏导）不一定可微，可导（偏导）也不一定连续

根据可导定义，令

lim⁡Δx→0ΔyΔx=A\lim\limits_ {\Delta x \to 0}\frac{\Delta y}{\Delta x} = AΔx→0lim​ΔxΔy​=A

则有

lim⁡Δx→0Δy−AΔxΔx=0\lim\limits_ {\Delta x \to 0}\frac{\Delta y - A\Delta x}{\Delta x} = 0Δx→0lim​ΔxΔy−AΔx​=0

即有Δy−AΔx=o(Δx)\Delta y - A\Delta x = o(\Delta x)Δy−AΔx=o(Δx)，故Δy=AΔ+o(Δx)\Delta y = A\Delta + o(\Delta x)Δy=AΔ+o(Δx)，其中AAA为常数，满足可微的定义，因此，可导必可微。

根据可微定义

Δy=AΔx+o(Δx)\Delta y = A\Delta x + o(\Delta x) Δy=AΔx+o(Δx)

则

f′(x0)=lim⁡Δx→0AΔx+o(Δx)Δx=Af'(x_0) = \lim\limits_{\Delta x \to 0}\frac{A \Delta x + o(\Delta x)}{\Delta x} = A f′(x0​)=Δx→0lim​ΔxAΔx+o(Δx)​=A

导数存在，故满足可导的定义，因此可微必可导，且f′(x)=Af'(x) = Af′(x)=A.

导数f′(x0)f'(x_0)f′(x0​)在几何上表示曲线y=f(x)y = f(x)y=f(x)在点(x0,f(x0))(x_0, f(x_0))(x0​,f(x0​))处切线的斜率。

注：法线的斜率是切线斜率的负倒数。

设x=x(t)x = x(t)x=x(t)及y=y(t)y = y(t)y=y(t)都是可导函数，而变量xxx与yyy之间存在某种关系，从而他们的变化率dxdt\dfrac{dx}{dt}dtdx​与dydt\dfrac{dy}{dt}dtdy​之间也存在一定关系，这样两个相互依赖的变化率成为相关变化率

已知动点PPP在曲线y=x3y = x^3y=x3上运动，记坐标原点与点PPP间的距离为lll。若点PPP的横坐标对时间的变化率为常数v0v_0v0​，则当点PPP运动到点(1,1)(1, 1)(1,1)时，lll对时间的变化率是‾\underline{\hspace*{1cm}}​.

解：

已知dxdv=v0\dfrac{dx}{dv} = v_0dvdx​=v0​，l=x2+x6l = \sqrt{x^2 + x^6}l=x2+x6​，则

dldt=dldx⋅dxdt=2x+6x52x2+x6⋅v0\frac{dl}{dt} = \frac{dl}{dx} \cdot \frac{dx}{dt} = \frac{2x + 6x^5}{2\sqrt{x^2 + x^6}} \cdot v_0 dtdl​=dxdl​⋅dtdx​=2x2+x6​2x+6x5​⋅v0​

带入数值x=1x = 1x=1，则

dldt=1+32v0=22v0\frac{dl}{dt} = \frac{1 + 3}{\sqrt{2}}v_0 = 2\sqrt{2} v_0 dtdl​=2​1+3​v0​=22​v0​

(C)′=0(2.1)(C)' = 0 \tag{2.1} (C)′=0(2.1)

(xa)′=axa−1(2.2)(x^a)' = ax^{a-1} \tag{2.2} (xa)′=axa−1(2.2)

(ax)′=axln⁡(a)(2.3)(a^x)' = a^x\ln(a) \tag{2.3} (ax)′=axln(a)(2.3)

(ex)′=ex(2.4)(e^x)' = e^x \tag{2.4} (ex)′=ex(2.4)

(log⁡ax)′=1xln⁡(a)(2.5)(\log_a^x)' = \frac{1}{x\ln(a)} \tag{2.5} (logax​)′=xln(a)1​(2.5)

(ln⁡∣x∣)′=1x(2.6)(\ln \mid x \mid )' = \frac{1}{x} \tag{2.6} (ln∣x∣)′=x1​(2.6)

(sin⁡x)′=cos⁡(x)(2.7)(\sin x)' = \cos(x) \tag{2.7} (sinx)′=cos(x)(2.7)

(cos⁡x)′=−sin⁡(x)(2.8)(\cos x)' = -\sin(x) \tag{2.8} (cosx)′=−sin(x)(2.8)

(tan⁡x)′=sec⁡2(x)(2.9)(\tan x )' = \sec^2(x) \tag{2.9} (tanx)′=sec2(x)(2.9)

(cot⁡x)′=−csc⁡2(x)(2.10)(\cot x)' = - \csc^2(x) \tag{2.10} (cotx)′=−csc2(x)(2.10)

(sec⁡x)′=sec⁡(x)tan⁡(x)(2.11)(\sec x)' = \sec (x) \tan (x) \tag{2.11} (secx)′=sec(x)tan(x)(2.11)

(csc⁡x)′=csc⁡2(x)cot⁡(x)(2.12)(\csc x)' = \csc^2(x) \cot (x) \tag{2.12} (cscx)′=csc2(x)cot(x)(2.12)

(arcsin⁡x)′=11−x2(2.13)(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}} \tag{2.13} (arcsinx)′=1−x2​1​(2.13)

(arccos⁡x)′=−11−x2(2.14)(\arccos x)' = - \frac{1}{\sqrt{1 - x^2}} \tag{2.14} (arccosx)′=−1−x2​1​(2.14)

(arctan⁡x)′=11+x2(2.15)(\arctan x)' = \frac{1}{1 + x^2} \tag{2.15} (arctanx)′=1+x21​(2.15)

(arcctg⁡x)′=11−x2(2.16)(\arcctg x)' = \frac{1}{\sqrt{1 - x^2}} \tag{2.16} (arcctgx)′=1−x2​1​(2.16)

注：sec⁡(x)=1cos⁡(x)\sec(x) = \dfrac{1}{\cos(x)}sec(x)=cos(x)1​，csc⁡(x)=1sin⁡(x)\csc(x) = \dfrac{1}{\sin(x)}csc(x)=sin(x)1​

设u=u(x),v=v(x)u = u(x), v = v(x)u=u(x),v=v(x)在xxx处可导，则

(u±v)′=u′±v′(2.17)(u \pm v)' = u' \pm v' \tag{2.17}(u±v)′=u′±v′(2.17)

(uv)′=u′v+uv′(2.18)(uv)' = u'v + uv' \tag{2.18}(uv)′=u′v+uv′(2.18)

(uv)′=u′v−uv′v2(2.19)(\dfrac{u}{v})' = \dfrac{u'v - uv'}{v^2} \tag{2.19}(vu​)′=v2u′v−uv′​(2.19)

设u=φ(x)u = \varphi(x)u=φ(x)在xxx处可导，y=f(u)y = f(u)y=f(u)在对应点可导，则复合函数y=f[φ(x)]y = f[\varphi(x)]y=f[φ(x)]在xxx处可导，则

dydx=dydu⋅dudx=f′(u)φ′(x)(2.20)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = f'(u)\varphi'(x) \tag{2.20} dxdy​=dudy​⋅dxdu​=f′(u)φ′(x)(2.20)

一个可导的奇（偶）函数，求一次导，其奇偶性发生一次变化

f(x)f(x)f(x)满足f(−x)=−f(x)f(-x) = -f(x)f(−x)=−f(x),又根据复合函数求导法则，得到f′(−x)=−f′(x)f'(-x) = -f'(x)f′(−x)=−f′(x)，则

[f(−x)]′=−[−f(x)]′=[f(x)]′[f(-x)]' = -[-f(x)]' = [f(x)]' [f(−x)]′=−[−f(x)]′=[f(x)]′

即f′(x)f'(x)f′(x)为偶函数

f(x)f(x)f(x)满足f(−x)=f(x)f(-x) = f(x)f(−x)=f(x),又根据复合函数求导法则，得到f′(−x)=−f′(x)f'(-x) = -f'(x)f′(−x)=−f′(x)，则

[f(−x)]′=−[f(x)]′[f(-x)]' = -[f(x)]' [f(−x)]′=−[f(x)]′

即f′(x)f'(x)f′(x)为奇函数

设y=y(x)y = y(x)y=y(x)是由方程F(x,y)=xF(x, y) = xF(x,y)=x所确定的可导函数，为求得y′y'y′，可在方程F(x,y)=0F(x, y) = 0F(x,y)=0两边对xxx求导，可得到一个含有y′y'y′的方程，从中解出y′y'y′即可。

注：y′y'y′也可由多元函数微分法中的隐函数求导公式2.21得到。

dydx=−Fx′Fy′(2.21)\frac{dy}{dx} = - \frac{F'_x}{F'_y} \tag{2.21} dxdy​=−Fy′​Fx′​​(2.21)

若y=f(x)y = f(x)y=f(x)在某区间内可导，且f′(x)≠0f'(x) \ne 0f′(x)​=0，则其反函数x=φ(x)x = \varphi (x)x=φ(x)在对应区间内也可导，且

φ(y)=1f′(x)(2.22)\varphi (y) = \frac{1}{f'(x)} \tag{2.22} φ(y)=f′(x)1​(2.22)

即

dydx=1dydx\frac{dy}{dx} =\frac{1}{\dfrac{dy}{dx}} dxdy​=dxdy​1​

设y=y(x)y = y(x)y=y(x)是由参数方程

{x=φ(x)y=ψ(x),(αx=ρsinθy=ρcosθ​(2.26)

已知经过点M(ρo,θ0)M(\rho_o, \theta_0)M(ρo​,θ0​)，且直线与极轴所成角为α\alphaα的直线lll，其极坐标方程为

ρsin⁡(α−θ)=ρ0sin⁡(α0−θ0)\rho \sin (\alpha - \theta) = \rho_0 \sin(\alpha_0 - \theta_0) ρsin(α−θ)=ρ0​sin(α0​−θ0​)

即

ρ=ρ0sec⁡(α0−θ0)\rho = \rho_0 \sec(\alpha_0 - \theta_0) ρ=ρ0​sec(α0​−θ0​)

转化为参数方程形式

{x=ρ0sec⁡(α0−θ0)sin⁡(θ)y=ρ0sec⁡(α0−θ0)cos⁡(θ){\left\{ \begin{aligned} x = \rho_0 \sec(\alpha_0 - \theta_0) \sin(\theta)\\ y = \rho_0 \sec(\alpha_0 - \theta_0) \cos(\theta)\\ \end{aligned} \right.} {x=ρ0​sec(α0​−θ0​)sin(θ)y=ρ0​sec(α0​−θ0​)cos(θ)​

如果y=y(x)y = y(x)y=y(x)的表达式由多个因式的乘除、乘幂构成，或是幂指函数的形式，则可先将函数去对数，然后两边对xxx求导。

注：对等式两边取对数，需要满足等式两边都大于0的条件

含义：一般地，函数y=f(x)y = f(x)y=f(x)的nnn阶导数为y(n)=[f(n−1)(x)]′y^{(n)} = [f^{(n - 1)}(x)]'y(n)=[f(n−1)(x)]′，也可记为f(n)(x)f^{(n)}(x)f(n)(x)或dnydxn\dfrac{d^ny}{dx^n}dxndny​，即nnn阶导数就是n−1n-1n−1阶导函数的导数。

注：如果函数在点xxx处nnn阶可导，则在点xxx的某邻域内f(x)f(x)f(x)必定具有一切低于nnn阶的导数。

(sin⁡x)(n)=sin⁡(x+n⋅π2)(2.27)(\sin x)^{(n)} = \sin (x + n \cdot \frac{\pi}{2}) \tag{2.27} (sinx)(n)=sin(x+n⋅2π​)(2.27)

(cosx)(n)=cos⁡(x+n⋅π2)(2.28)(cos x)^{(n)} = \cos (x + n \cdot \frac{\pi}{2}) \tag{2.28} (cosx)(n)=cos(x+n⋅2π​)(2.28)

(u±v)(n)=u(n)±v(n)(2.29)(u \pm v)^{(n)} = u^{(n)} \pm v^{(n)} \tag{2.29} (u±v)(n)=u(n)±v(n)(2.29)

(uv)(n)=∑k=0nCnku(k)v(n−k)(2.30)(uv)^{(n)} = \sum_{k=0}^n C_n^k u^{(k)}v^{(n-k)} \tag{2.30} (uv)(n)=k=0∑n​Cnk​u(k)v(n−k)(2.30)

式2.24可类比nnn阶二项式公式

(u+v)n=∑k=0nCnkukvn−k(2.31)(u + v)^{n} = \sum_{k=0}^n C_n^k u^{k}v^{n-k} \tag{2.31} (u+v)n=k=0∑n​Cnk​ukvn−k(2.31)

若y=sin⁡(ax+b)y= \sin(ax + b)y=sin(ax+b)，则 y(n)=ansin⁡(ax+b+n⋅π2)(2.32)y^{(n)} = a^n \sin(ax + b + n \cdot \frac{\pi}{2}) \tag{2.32} y(n)=ansin(ax+b+n⋅2π​)(2.32)

通过归纳法，求y′y'y′和y′′y''y′′，推出y(n)y^{(n)}y(n).