高等数学期末总复习 DAY 3.利用导数定义求极限 判断连续与可导的关系 关于导数定义的证明题 基本求导 基本高阶求导 抽象函数求导 |
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DAY 3.
一路陪我走过来的从来都不是什么善良正直正能量,而是虚荣嫉妒不甘心 文章目录 DAY 3.1. 利用导数定义求极限2.判断连续与可导的关系3.关于导数定义的证明题4.基本复合函数求导5. 基本高阶求导6. 抽象函数求导 1. 利用导数定义求极限导数的两种定义 f ′ ( x 0 ) f'{(x_0)} f′(x0) = lim x → x 0 f ( x ) − f ( x 0 ) x − x 0 \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} limx→x0x−x0f(x)−f(x0) f ′ ( x 0 ) f'{(x_0)} f′(x0) = lim Δ x → 0 \lim_{ \Delta x \to 0} limΔx→0 f ( x 0 + Δ x ) − f ( x 0 ) Δ x \frac{f(x_0 + \Delta x) - f(x_0)}{ \Delta x} Δxf(x0+Δx)−f(x0)解题基本用到的是凑上面两种形式的思想。 例题1 求 A = lim Δ x → 0 \lim_{ \Delta x \to 0} limΔx→0 f ( x − Δ x ) − f ( x ) Δ x \frac{f(x - \Delta x) - f(x)}{ \Delta x} Δxf(x−Δx)−f(x) 的极限 这里和上面第二类导数的定义就相差一个正负号 f ( x + Δ x ) f(x + \Delta x) f(x+Δx)所以很显然我们要凑第二类导数的定义 解: A = lim Δ x → 0 \lim_{ \Delta x \to 0} limΔx→0 f ( x 0 − Δ x ) − f ( x 0 ) Δ x \frac{f(x_0 - \Delta x) - f(x_0)}{ \Delta x} Δxf(x0−Δx)−f(x0) = lim Δ x → 0 \lim_{ \Delta x \to 0} limΔx→0 f ( x 0 + ( − Δ x ) ) − f ( x 0 ) − Δ x \frac{f(x_0 +(- \Delta x)) - f(x_0)}{ -\Delta x} −Δxf(x0+(−Δx))−f(x0) *(-1) = f ′ ( x 0 ) f'(x_0) f′(x0) * (-1) = - f ′ ( x 0 ) f'(x_0) f′(x0) 例题2 如果 f ( 0 ) = 0 f(0) = 0 f(0)=0 ,且 f ′ ( 0 ) f'(0) f′(0)存在,求A = lim x → 0 f ( x ) x \lim_{x \to 0} \frac{f(x)}{x} limx→0xf(x) 解:由题意得 A = lim x → 0 f ( x ) − 0 x − 0 \lim_{x \to 0} \frac{f(x) - 0}{x - 0} limx→0x−0f(x)−0 = lim x → 0 f ( x ) − f ( 0 ) x − 0 \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} limx→0x−0f(x)−f(0) = f ′ ( 0 ) f'(0) f′(0) 例题3 求A = lim h → 0 f ( x 0 + h ) − f ( x 0 − h ) h \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0 - h)}{h} limh→0hf(x0+h)−f(x0−h) 解:原式 = lim h → 0 f ( x 0 + h ) − f ( x 0 ) + f ( x 0 ) − f ( x 0 − h ) h \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0) +f(x_0)- f(x_0 - h)}{h} limh→0hf(x0+h)−f(x0)+f(x0)−f(x0−h) = lim h → 0 f ( x 0 + h ) − f ( x 0 ) h \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0) }{h} limh→0hf(x0+h)−f(x0) - lim h → 0 f ( x 0 − h ) − f ( x 0 ) h \lim_{h \to 0} \frac{f(x_0 - h) - f(x_0) }{h} limh→0hf(x0−h)−f(x0) = f ′ ( x 0 ) f'(x_0) f′(x0) - lim h → 0 f ( x 0 − h ) − f ( x 0 ) h \lim_{h \to 0} \frac{f(x_0 - h) - f(x_0) }{h} limh→0hf(x0−h)−f(x0) = f ′ ( x 0 ) f'(x_0) f′(x0) - (- f ′ ( x 0 ) f'(x_0) f′(x0)) =2 f ′ ( x 0 ) f'(x_0) f′(x0) 2.判断连续与可导的关系可导一定连续,连续不一定可导 一般有两种题型: f ( x ) = { . . . ( x ! = x 0 ) . . . ( x = x 0 ) f(x)=\left\{ \begin{aligned} ... & & (x != x_0) \\ ... & & (x = x_0) \\ \end{aligned} \right. f(x)={......(x!=x0)(x=x0)例题4 讨论 f ( x ) = { 0 x=0 x 2 s i n 1 x x!=0 f(x)=\begin{cases} 0& \text{x=0}\\x^2sin\frac{1}{x}& \text{x!=0} \end{cases} f(x)={0x2sinx1x=0x!=0 的连续性与可导性 解:依题意得该函数的断点为 x = 0 则: lim x → 0 x 2 sin 1 x \lim_{x \to 0} x^2 \sin \frac{1}{x} limx→0x2sinx1 = 0 (DAY 1.中的一个重要结论 无穷小量*有界函数 = 0) 由此可知该函数连续。 而: f ′ ( 0 ) f'(0) f′(0) = lim x → 0 f ( x ) − f ( 0 ) x − 0 \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} limx→0x−0f(x)−f(0) = lim x → 0 x 2 sin 1 x − 0 x − 0 \lim_{x \to 0} \frac{x^2\sin \frac{1}{x} - 0}{x - 0} limx→0x−0x2sinx1−0 = 0 所以该函数也可导 f ( x ) = { . . . ( x ≤ x 0 ) . . . ( x > x 0 ) f(x)=\left\{ \begin{aligned} ... & & (x \le x_0) \\ ... & & (x > x_0) \\ \end{aligned} \right. f(x)={......(x≤x0)(x>x0)例题5 f ( x ) = { x 2 x 1 f(x)=\begin{cases} x^2& \text{x 1} \end{cases} f(x)={x2ax+bx 1 在 x = 1处可导,求a,b 解:首先函数可导则一定连续 可得, lim x → 1 a x 2 + b = f ( 1 ) = 1 \lim_{x \to 1} ax^2 + b = f(1) = 1 limx→1ax2+b=f(1)=1 可推 ⇒ \Rightarrow ⇒ a + b = 1 然后,函数可导,则左右导数相等; 利用定义求导可得, f ′ ( 1 − ) = lim x → 1 − f ( x ) − f ( 1 ) x − 1 f'(1^-) = \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} f′(1−)=limx→1−x−1f(x)−f(1) = lim x → 1 − x 2 − 1 x − 1 \lim_{x \to 1^-} \frac{x^2 - 1}{x - 1} limx→1−x−1x2−1 = 2 f ′ ( 1 + ) = lim x → 1 + f ( x ) − f ( 1 ) x − 1 f'(1^+) = \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} f′(1+)=limx→1+x−1f(x)−f(1) = lim x → 1 + a x + b − 1 x − 1 \lim_{x \to 1^+} \frac{ax +b - 1}{x - 1} limx→1+x−1ax+b−1 = lim x → 1 + ( 1 − b ) x + b − 1 x − 1 \lim_{x \to 1^+} \frac{(1-b)x +b - 1}{x - 1} limx→1+x−1(1−b)x+b−1 = lim x → 1 + ( 1 − b ) ( x − 1 ) x − 1 \lim_{x \to 1^+} \frac{(1-b)(x - 1)}{x - 1} limx→1+x−1(1−b)(x−1) = 1-b 所以: 1 - b = 2 ⇒ \Rightarrow ⇒ b = -1 由于:a + b = 1 ⇒ \Rightarrow ⇒ a = 2 3.关于导数定义的证明题例题6 设 f (x) 满足条件: f ( x + y ) = f ( x ) f ( y ) ; x , y ∈ R f(x + y) = f(x)f(y) ; x,y \in R f(x+y)=f(x)f(y);x,y∈R f ( x ) = 1 + x g ( x ) , lim x → 0 g ( x ) = 1 f(x) = 1+xg(x), \lim_{x \to 0} g(x) = 1 f(x)=1+xg(x),limx→0g(x)=1 证明f(x)在 R 上处处可导,且 f ′ ( x ) = f ( x ) f'(x) = f(x) f′(x)=f(x)解: f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} f′(x)=limΔx→0Δxf(x+Δx)−f(x) = lim Δ x → 0 f ( x ) f ( Δ x ) − f ( x ) Δ x \lim_{\Delta x \to 0} \frac{f(x)f( \Delta x) - f(x)}{\Delta x} limΔx→0Δxf(x)f(Δx)−f(x) = lim Δ x → 0 f ( x ) ( f ( Δ x ) − 1 ) Δ x \lim_{\Delta x \to 0} \frac{f(x)(f( \Delta x) - 1)}{\Delta x} limΔx→0Δxf(x)(f(Δx)−1) = f ( x ) lim Δ x → 0 f ( Δ x ) − 1 Δ x f(x)\lim_{\Delta x \to 0} \frac{f( \Delta x) - 1}{\Delta x} f(x)limΔx→0Δxf(Δx)−1 = f ( x ) lim Δ x → 0 1 + Δ x g ( Δ x ) − 1 Δ x f(x)\lim_{\Delta x \to 0} \frac{1 + \Delta x g(\Delta x)- 1}{\Delta x} f(x)limΔx→0Δx1+Δxg(Δx)−1 = f ( x ) lim Δ x → 0 g ( Δ x ) f(x)\lim_{\Delta x \to 0} g(\Delta x) f(x)limΔx→0g(Δx) = f ( x ) f(x) f(x) 证毕 4.基本复合函数求导注意牢记基本公式 5. 基本高阶求导和例题7安排在一起 注意求导的先后次序,以及中间是否可以化简等,不骜述。 6. 抽象函数求导例题7 包含第五点基本高阶求导 求 y = f ( x 2 ) y = f(x^2) y=f(x2) 的 d y d x , d 2 y d x 2 \frac{d_y}{d_x}, \frac{d^2{_y}}{d_x^2} dxdy,dx2d2y 解: d y d x \frac{d_y}{d_x} dxdy = f ′ ( x 2 ) ∗ 2 x f'(x^2) * 2x f′(x2)∗2x (先函数求导再中间量求导) d 2 y d x 2 \frac{d^2{_y}}{d_x^2} dx2d2y = ( f ′ ( x 2 ) ∗ 2 x ) ′ (f'(x^2) * 2x)' (f′(x2)∗2x)′ (乘积的求导) = f ′ ′ ( x 2 ) ∗ 2 x ∗ 2 x + f ′ ( x 2 ) ∗ 2 f''(x^2) *2x *2x + f'(x^2) *2 f′′(x2)∗2x∗2x+f′(x2)∗2 = 4 x 2 f ′ ′ ( x 2 ) + 2 f ′ ( x 2 ) 4x^2 f''(x^2) + 2f'(x^2) 4x2f′′(x2)+2f′(x2) |
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