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)
充分必要
条件注
: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 Δ x → 0 Δ y Δ x = A \lim\limits_ {\Delta x \to 0}\frac{\Delta y}{\Delta x} = A Δx→0limΔxΔy=A
则有
lim Δ x → 0 Δ y − A Δ x Δ x = 0 \lim\limits_ {\Delta x \to 0}\frac{\Delta y - A\Delta x}{\Delta x} = 0 Δx→0limΔxΔy−AΔx=0
即有 Δ y − A Δ x = o ( Δ x ) \Delta y - A\Delta x = o(\Delta x) Δy−AΔx=o(Δx),故 Δ y = A Δ + o ( Δ x ) \Delta y = A\Delta + o(\Delta x) Δy=AΔ+o(Δx),其中 A A A为常数,满足可微的定义,因此,可导必可微。
根据可微定义
Δ y = A Δ x + o ( Δ x ) \Delta y = A\Delta x + o(\Delta x) Δy=AΔx+o(Δx)
则
f ′ ( x 0 ) = lim Δ x → 0 A Δ x + o ( Δ x ) Δ x = A f'(x_0) = \lim\limits_{\Delta x \to 0}\frac{A \Delta x + o(\Delta x)}{\Delta x} = A f′(x0)=Δx→0limΔxAΔx+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 ) = A f'(x) = A f′(x)=A.
不能
推出 f ′ ( x ) f'(x) f′(x)在 x 0 x_0 x0点连续不能
推出 lim x → x 0 f ′ ( x ) \lim\limits_{x \to x_0}f'(x) x→x0limf′(x)存在导数 f ′ ( x 0 ) f'(x_0) f′(x0)在几何上表示曲线 y = f ( x ) y = f(x) y=f(x)在点 ( x 0 , f ( x 0 ) ) (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 x x x与 y y y之间存在某种关系,从而他们的变化率 d x d t \dfrac{dx}{dt} dtdx与 d y d t \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 P P P在曲线 y = x 3 y = x^3 y=x3上运动,记坐标原点与点 P P P间的距离为 l l l。若点 P P P的横坐标对时间的变化率为常数 v 0 v_0 v0,则当点 P P P运动到点 ( 1 , 1 ) (1, 1) (1,1)时, l l l对时间的变化率是 ‾ \underline{\hspace*{1cm}} .
解:
已知 d x d v = v 0 \dfrac{dx}{dv} = v_0 dvdx=v0, l = x 2 + x 6 l = \sqrt{x^2 + x^6} l=x2+x6,则
d l d t = d l d x ⋅ d x d t = 2 x + 6 x 5 2 x 2 + x 6 ⋅ v 0 \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+x62x+6x5⋅v0
带入数值 x = 1 x = 1 x=1,则
d l d t = 1 + 3 2 v 0 = 2 2 v 0 \frac{dl}{dt} = \frac{1 + 3}{\sqrt{2}}v_0 = 2\sqrt{2} v_0 dtdl=21+3v0=22v0
( C ) ′ = 0 (2.1) (C)' = 0 \tag{2.1} (C)′=0(2.1)
( x a ) ′ = a x a − 1 (2.2) (x^a)' = ax^{a-1} \tag{2.2} (xa)′=axa−1(2.2)
( a x ) ′ = a x ln ( a ) (2.3) (a^x)' = a^x\ln(a) \tag{2.3} (ax)′=axln(a)(2.3)
( e x ) ′ = e x (2.4) (e^x)' = e^x \tag{2.4} (ex)′=ex(2.4)
( log a x ) ′ = 1 x ln ( a ) (2.5) (\log_a^x)' = \frac{1}{x\ln(a)} \tag{2.5} (logax)′=xln(a)1(2.5)
( ln ∣ x ∣ ) ′ = 1 x (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 ) ′ = 1 1 − x 2 (2.13) (\arcsin x)' = \frac{1}{\sqrt{1 - x^2}} \tag{2.13} (arcsinx)′=1−x21(2.13)
( arccos x ) ′ = − 1 1 − x 2 (2.14) (\arccos x)' = - \frac{1}{\sqrt{1 - x^2}} \tag{2.14} (arccosx)′=−1−x21(2.14)
( arctan x ) ′ = 1 1 + x 2 (2.15) (\arctan x)' = \frac{1}{1 + x^2} \tag{2.15} (arctanx)′=1+x21(2.15)
( arcctg x ) ′ = 1 1 − x 2 (2.16) (\arcctg x)' = \frac{1}{\sqrt{1 - x^2}} \tag{2.16} (arcctgx)′=1−x21(2.16)
注
: sec ( x ) = 1 cos ( x ) \sec(x) = \dfrac{1}{\cos(x)} sec(x)=cos(x)1, csc ( x ) = 1 sin ( 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 x x x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)
( u v ) ′ = u ′ v + u v ′ (2.18) (uv)' = u'v + uv' \tag{2.18} (uv)′=u′v+uv′(2.18)
( u v ) ′ = u ′ v − u v ′ v 2 (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 x x x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 x x x处可导,则
d y d x = d y d u ⋅ d u d x = 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
奇函数
。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)]' [f(−x)]′=−[−f(x)]′=[f(x)]′
即 f ′ ( x ) f'(x) f′(x)为偶函数
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 ) = x F(x, y) = x F(x,y)=x所确定的可导函数,为求得 y ′ y' y′,可在方程 F ( x , y ) = 0 F(x, y) = 0 F(x,y)=0两边对 x x x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。
d y d x = − F x ′ F y ′ (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 ) ≠ 0 f'(x) \ne 0 f′(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 ) = 1 f ′ ( x ) (2.22) \varphi (y) = \frac{1}{f'(x)} \tag{2.22} φ(y)=f′(x)1(2.22)
即
d y d x = 1 d y d x \frac{dy}{dx} =\frac{1}{\dfrac{dy}{dx}} dxdy=dxdy1
设 y = y ( x ) y = y(x) y=y(x)是由参数方程
{ x = φ ( x ) y = ψ ( x ) , ( α < t < β ) {\left\{ \begin{aligned} &x = \varphi (x)\\ &y = \psi (x)\\ \end{aligned}\right. }, (\alpha < t < \beta) {x=φ(x)y=ψ(x),(α<t<β)
确定的函数,则
d y d x = ψ ( x ) φ ( x ) (2.23) \frac{dy}{dx} = \frac{\psi(x)}{\varphi(x)} \tag{2.23} dxdy=φ(x)ψ(x)(2.23)
d 2 y d 2 x = d d t ( d y d x ) ⋅ d t d x = d d t ( ψ ′ ( t ) φ ′ ( t ) ) ⋅ 1 φ ′ ( x ) = ψ ′ ′ ( t ) φ ′ ( x ) − φ ′ ′ ( x ) ψ ′ ( t ) φ 3 ( t ) (2.24) \frac{d^2 y}{d^2 x} = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx}= \frac{d}{dt}(\frac{\psi '(t)}{\varphi '(t)}) \cdot \frac{1}{\varphi '(x)} = \frac{\psi ''(t)\varphi '(x) - \varphi ''(x) \psi '(t)}{\varphi^3 (t)} \tag{2.24} d2xd2y=dtd(dxdy)⋅dxdt=dtd(φ′(t)ψ′(t))⋅φ′(x)1=φ3(t)ψ′′(t)φ′(x)−φ′′(x)ψ′(t)(2.24)
极坐标性质
{ ρ 2 = x 2 + y 2 tan θ = y x ( x ≠ 0 ) (2.25) {\left\{ \begin{aligned} \rho^2 &= x^2 + y^2\\ \tan \theta &= \frac{y}{x} (x \ne 0)\\ \end{aligned}\right.} \tag{2.25} ⎩⎨⎧ρ2tanθ=x2+y2=xy(x=0)(2.25)
极坐标转化为直角坐标的转化公式
{ x = ρ sin θ y = ρ cos θ (2.26) {\left\{ \begin{aligned} x = \rho \sin \theta\\ y = \rho \cos \theta\\ \end{aligned}\right.} \tag{2.26} {x=ρsinθy=ρcosθ(2.26)
已知经过点 M ( ρ o , θ 0 ) M(\rho_o, \theta_0) M(ρo,θ0),且直线与极轴所成角为 α \alpha α的直线 l l l,其极坐标方程为
ρ sin ( α − θ ) = ρ 0 sin ( α 0 − θ 0 ) \rho \sin (\alpha - \theta) = \rho_0 \sin(\alpha_0 - \theta_0) ρsin(α−θ)=ρ0sin(α0−θ0)
即
ρ = ρ 0 sec ( α 0 − θ 0 ) \rho = \rho_0 \sec(\alpha_0 - \theta_0) ρ=ρ0sec(α0−θ0)
转化为参数方程形式
{ x = ρ 0 sec ( α 0 − θ 0 ) sin ( θ ) y = ρ 0 sec ( α 0 − θ 0 ) 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=ρ0sec(α0−θ0)sin(θ)y=ρ0sec(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 x x x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 n n n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 d n y d x n \dfrac{d^ny}{dx^n} dxndny,即 n n n阶导数就是 n − 1 n-1 n−1阶导函数的导数。
注
:如果函数在点 x x x处 n n n阶可导,则在点 x x x的某邻域内 f ( x ) f(x) f(x)必定具有一切低于 n n n阶的导数。
( sin x ) ( n ) = sin ( x + n ⋅ π 2 ) (2.27) (\sin x)^{(n)} = \sin (x + n \cdot \frac{\pi}{2}) \tag{2.27} (sinx)(n)=sin(x+n⋅2π)(2.27)
( c o s x ) ( n ) = cos ( x + n ⋅ π 2 ) (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)
( u v ) ( n ) = ∑ k = 0 n C n k u ( 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∑nCnku(k)v(n−k)(2.30)
式2.24可类比 n n n阶二项式公式
( u + v ) n = ∑ k = 0 n C n k u k v n − 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∑nCnkukvn−k(2.31)
若 y = sin ( a x + b ) y= \sin(ax + b) y=sin(ax+b),则
y ( n ) = a n sin ( a x + b + n ⋅ π 2 ) (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).