高数 02.02函数的求导法则

第二章第二节函数的求导法则 \color{blue}{第二章第二节函数的求导法则}

一、四则运算的求导法则
二、复合函数的求导法则
三、初等函数的求导问题

思路：
f ′ (x)=lim Δx→0 f(x+Δx)−f(x)Δx (构造性定义) f^{\prime}(x) = \lim_{\Delta {x} \rightarrow 0}{\dfrac{f(x + \Delta{x}) - f(x)}{\Delta{x}}}(构造性定义)
⇓   ⇓ \qquad \Downarrow\qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ \Downarrow
  求导法则⇓   \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \left. \begin{array}{c}{\color{green}{求导法则}} \\ \Downarrow \end{array} \right.
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (C) ′ =0(sinx) ′ =cosx(lnx) ′ =1x  ⎫ ⎭ ⎬ 证明中利用了两个重要极限  其他基本初等函数求导公式                                                          初等函数求导问题   \begin{matrix} \underbrace{ \left \{ \begin{array}{l}(C)^{\prime} = 0 \\ \left. \begin{array}{l} (\sin{x})^{\prime} = cos{x} \\ (\ln{x})^{\prime} = \dfrac{1}{x} \end{array} \right \} \left. \begin{array}{l}证明中利用了 \\ 两个重要极限 \end{array} \right. \end{array} \right. \quad {\color{green}{\left. \begin{array}{l} 其他基本初等\\ 函数求导公式 \end{array} \right.}}} \\ {\color{green}{初等函数求导问题}}\end{matrix}

一、四则运算的求导法则 \color{blue}{一、四则运算的求导法则}

定理1.函数u=u(x)及v=v(x)都在x具有导数⟹u(x)及v(x)的和、差、积、商(除分母为0的点外)都在点x可导,且(1)[u(x)±v(x)] ′ =u ′ (x)±v ′ (x)(2)[u(x)v(x)] ′ =u ′ (x)v(x)+u(x)v ′ (x)(3)[u(x)v(x) ] ′ =u ′ (x)v(x)−u(x)v ′ (x)v 2 x (v(x)≠0)用导数的定义可以证明上面的3个公式定理1.函数u = u(x) 及 v = v(x)都在x具有导数 \\ \Longrightarrow u(x) 及 v(x)的和、差、积、商(除分母为0的点外)都在点x可导,且 \\ (1)[u(x) \pm v(x)]^{\prime} = u^{\prime}(x) \pm v^{\prime}(x) \\ (2)[u(x)v(x)]^{\prime} = u^{\prime}(x) v(x) + u(x)v^{\prime}(x) \\ (3)[\dfrac{u(x)}{v(x)}]^{\prime} = \dfrac{u^{\prime}(x)v(x) - u(x)v^{\prime}(x)}{v^2{x}} \quad (v(x) \neq 0) \\ 用导数的定义可以证明上面的3个公式
推论：1)(Cu) ′ =Cu ′ (C为常数)2)(uvw) ′ =u ′ vw+uv ′ w+uvw ′ 3)(log a x) ′ =(lnxlna ) ′ =1xlna 4)(Cv ) ′ =−Cv ′ v 2 (C为常数)5)(u+v−w) ′ =u ′ +v ′ −w ′ 推论：\\ 1) (Cu)^{\prime} = Cu^{\prime} (C为常数) \\ 2)(uvw)^{\prime} = u^{\prime}vw + uv^{\prime}w + uvw^{\prime} \\ 3)(log_a{x})^{\prime} = \big( \dfrac{\ln {x}}{\ln {a}} \big)^{\prime} = \dfrac{1}{x \ln{a}} \\ 4)(\dfrac{C}{v})^{\prime} = \dfrac{-Cv^{\prime}}{v^2}(C为常数) \\ 5)(u + v -w)^{\prime} = u^{\prime} + v^{\prime} - w^{\prime}

例1.y=x √ (x 3 −4cosx−sin1),求y ′ 及y ′ ∣ ∣ x=1 . 例1. y = \sqrt{x}(x^3 - 4\cos{x} - sin{1}), 求y^{\prime}及\left. y^{\prime} \right|_{x = 1} .
解：y ′ =12 x 3 −4cosx−sin1x √ +x √ (3x 2 +4sinx−0)=7x 3 +8xsinx−4cosx−sin12x √ y ′ ∣ ∣ x=1 =7+7sin1−4cos12 =72 +72 sin1−2cos1 \color{blue}{解：\\ y^{\prime} = \dfrac{1}{2}\dfrac{x^3 - 4\cos{x} - sin{1}}{\sqrt{x}} + \sqrt{x} (3x^2 + 4\sin{x} - 0) = \dfrac{7x^3 +8x \sin{x} -4\cos{x} - \sin1}{2\sqrt{x}} \\ \left. y^{\prime} \right|_{x = 1} \\ = \dfrac{7 + 7\sin{1} - 4\cos{1}}{2} \\ = \dfrac{7}{2} + \dfrac{7}{2}{sin{1}} - 2\cos{1} }

例2.求证(tanx) ′ =sec 2 x,(cscx) ′ =−cscxcotx. 例2.求证(\tan{x})^{\prime} = \sec^2{x}, (\csc{x})^{\prime} = -\csc{x} \cot{x} .
证：(tanx) ′ =(sinxcosx ) ′ =sin ′ xcosx−sinxcos ′ xcos 2 x =cos 2 x+sin 2 xcos 2 x =1cos 2 x =sec 2 x (cscx) ′ =(1sinx ) ′ =−(sinx) ′ sin 2 x =−cosxsinxsinx =−cscxcotx 类似可证：(cotx) ′ =−csc 2 x,(secx) ′ =secxtanx \color{blue}{证：\\ (\tan{x})^{\prime} = \Big( \dfrac{\sin{x}}{\cos{x}} \Big)^{\prime} \\ = \dfrac{\sin^{\prime}{x} \cos{x} - \sin{x} \cos^{\prime}{x}}{\cos^2{x}} \\ = \dfrac{\cos^2{x} + \sin^2{x}}{\cos^2{x}} \\ = \dfrac{1}{\cos^2{x}} \\ = \sec^2{x} \\ \ \\\ (\csc{x})^{\prime} = \Big( \dfrac{1}{\sin{x}} \Big)^{\prime} \\ = \dfrac{-(\sin{x})^{\prime}}{\sin^2{x}} \\ = -\dfrac{\cos{x}}{\sin{x} \sin{x}} \\ = -\csc{x} \cot{x} \\ \ \\ 类似可证：\\ (\cot{x})^{\prime} = -\csc^2{x}, \\ (sec{x})^{\prime} = \sec{x} \tan{x} }

二、复合函数的求导法则 \color{blue}{二、复合函数的求导法则}

定理2.u=g(x)在点x可导,y=f(u)在点u=g(x)可导⟹复合函数y=f[g(x)]在点x可导，且dydx =f ′ (u)g ′ (x) 定理2.u=g(x)在点x可导, y = f(u)在点u=g(x)可导 \\ \Longrightarrow 复合函数y= f[g(x)]在点x可导，且\\ \dfrac{dy}{dx} = f^{\prime}(u)g^{\prime}(x)
证：∵y=f(u)在点u可导，故lim Δu→0 ΔyΔx =f ′ (u)即ΔyΔu =f ′ (u)+α∴Δy=f ′ Δu+αΔu(当Δu→0时α→0)故有ΔyΔx =f ′ (u)ΔuΔx +αΔuΔx (Δx≠0)∴dydx =lim Δx→0 ΔyΔx =lim Δx→0 [f ′ (u)ΔuΔx +αΔuΔx ]=f ′ (u)g ′ (x) \color{blue}{证： \\ \because y = f(u)在点u可导，故\lim_{\Delta{u} \rightarrow 0}{\dfrac{\Delta{y}}{\Delta{x}}} = f^{\prime}(u) \\ 即 \dfrac{\Delta{y}}{\Delta{u}} = f^{\prime}(u) + \alpha \\ \therefore \Delta{y} = f^{\prime}{\Delta{u}} + \alpha {\Delta{u}} (当\Delta{u} \rightarrow 0时\alpha \rightarrow 0) \\ 故有 \quad \dfrac{\Delta{y}}{\Delta{x}} = f^{\prime}(u) \dfrac{\Delta{u}}{\Delta{x}} + \alpha \dfrac{\Delta{u}}{\Delta{x}} (\Delta{x} \neq 0) \\ \therefore \dfrac{dy}{dx} = \lim_{\Delta{x} \rightarrow 0}{\dfrac{\Delta{y}}{\Delta{x}}} = \lim_{\Delta{x} \rightarrow 0}{[f^{\prime}(u)\dfrac{\Delta{u}}{\Delta{x}} + \alpha \dfrac{\Delta{u}}{\Delta{x}} ] } \\ = f^{\prime}(u)g^{\prime}(x) }
推广：此法则可推广到多个中间变量的情形推广：此法则可推广到多个中间变量的情形
例如：y=f(u),u=φ(v),v=ψ(x)dydx =dydu ⋅dudv ⋅dvdx =f ′ (u)⋅φ ′ (v)⋅ψ ′ (x) 例如：y = f(u), u = \varphi(v), v = \psi(x) \\ \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dv} \cdot \dfrac{dv}{dx} \\ = f^{\prime}(u) \cdot \varphi^{\prime}(v) \cdot \psi^{\prime}(x)

例3.求下列导数：(1)(x μ ) ′ ;(2)(x x ) ′ ;(3)(sinhx) ′ . 例3.求下列导数：(1)(x^{\mu})^{\prime};(2) (x^x)^{\prime}; (3)(\sinh {x})^{\prime} .
解：(1)(x μ ) ′ =(e μlnx ) ′ =e μlnx ⋅(μlnx)′=x μ ⋅μx =μx μ−1 (2)(x x ) ′ =(e xlnx ) ′ =(e xlnx ) ′ ⋅(xlnx) ′ =x x (lnx+1)(3)(sinhx) ′ =(e x −e −x 2 ) ′ =e x +e −x 2 =coshx说明：类似可得(coshx) ′ =sinhx;(tanhx) ′ =1cosh 2 x ;(a x ) ′ =e xlna \color{blue}{解： (1)(x^{\mu})^{\prime} = (e^{\mu \ln{x}})^{\prime} = e^{\mu \ln{x}} \cdot (\mu \ln{x}){\prime} = x^{\mu} \cdot \dfrac{\mu}{x} = \mu x^{\mu - 1} \\ (2) (x^x)^{\prime} = (e^{x \ln{x}})^{\prime} = (e^{x \ln{x}})^{\prime} \cdot (x \ln{x})^{\prime} \\ = x^x(\ln{x} + 1) \\ (3) (\sinh {x})^{\prime} = (\dfrac{e^{x} - e^{-x}}{2})^{\prime} = \dfrac{e^{x} + e^{-x}}{2} = \cosh{x} \\ 说明：类似可得 \\ (cosh{x})^{\prime} = \sinh{x}; (tanh{x})^{\prime} = \dfrac{1}{cosh^2{x}};(a^x)^{\prime} = e^{x \ln{a}} }

例4.设y=lncos(e x ),求dydx . 例4.设 y = \ln{\cos(e^x)},求\dfrac{dy}{dx}.
解：dydx =1cos(e x ) ⋅(−sin(e x ))⋅e x =−e x tan(e x ) \color{blue}{解：\\ \dfrac{dy}{dx} = \dfrac{1}{cos(e^x)} \cdot (-sin(e^x)) \cdot e^x \\ = -e^x \tan(e^x) }

三、初等函数的求导问题 \color{blue}{三、初等函数的求导问题}

1.常数和基本初等函数的导数(P94)
(C) ′ =0(x μ ) ′ =μx μ−1 (sinx) ′ =cosx(cosx) ′ =−sinx(tanx) ′ =sec 2 x(cotx) ′ =−csc 2 x(secx) ′ =secxtanx(cscx) ′ =−cscxcotx(a x ) ′ =a x lna(e x ) ′ =e x (log a x) ′ =1xlna (lnx) ′ =1x \color{blue}{ (C)^{\prime} = 0 \quad (x^{\mu})^{\prime} = \mu x^{\mu - 1} \\ (\sin{x})^{\prime} = \cos{x} \quad (\cos{x})^{\prime} = -\sin{x} \\ (\tan{x})^{\prime} = \sec^2{x} \quad (\cot{x})^{\prime} = -\csc^2{x} \\ (\sec{x})^{\prime} = \sec{x}\tan{x} \quad (\csc{x})^{\prime} = -csc{x} \cot{x} \\ (a^x)^{\prime} = a^x \ln{a} \quad (e^x)^{\prime} = e^x \\ (\log_a{x})^{\prime} = \dfrac{1}{x \ln{a}} \quad (\ln{x})^{\prime} = \dfrac{1}{x} \\ }

2.有限次四则运算的求导法则
(u±v) ′ =u ′ ±v ′ (Cu) ′ =Cu ′ (uv) ′ =u ′ v+uv ′ (uv ) ′ =u ′ v−uv ′ v 2 (v≠0) \color{blue}{(u \pm v)^{\prime} = u^{\prime} \pm v^{\prime} \quad (Cu)^{\prime} = Cu^{\prime} \\ (uv)^{\prime} = u^{\prime}v + uv^{\prime} \quad (\dfrac{u}{v})^{\prime} = \dfrac{u^{\prime}v - uv^{\prime}}{v^2} \quad (v \neq 0) }

3.复合函数的求导法则
y=f(u),u=φ(x)dydx =dydu ⋅dudx =f ′ (u)⋅φ ′ (x) \color{blue}{ y = f(u), u = \varphi(x) \\ \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} = f^{\prime}(u) \cdot \varphi^{\prime}(x) }

4.初等函数在定义域区间内可导，且导数仍为初等函数

说明：最基本的公式：
(C) ′ =0(sinx) ′ =cosx(lnx) ′ =1x (C)^{\prime} = 0 \\ (\sin{x})^{\prime} = \cos{x} \\ (\ln{x})^{\prime} = \dfrac{1}{x}

内容小结：
求导公式及求导法则(P94)
注意:
1)(uv) ′ ≠u ′ v ′ ,(uv ) ′ ≠u ′ v ′ 1)(uv)^{\prime} \neq u^{\prime}v^{\prime}, \quad (\dfrac{u}{v})^{\prime} \neq \dfrac{u^{\prime}}{v^{\prime}}
2)搞清楚复合函数结构，有外向内逐层求导.