节: 高阶导数 | 3.11高阶导数

章节大纲

You may recall hearing about Becca and her Track and Field competition in a prior concept. Her boyfriend had been recording on her laptop the position signals from the GPS transmitter Becca was wearing during her race. She was using a laptop program she had written to process the once per second GPS position signals so she could determine how far she had run at time during the race. These distance and time points were plotted on her laptop. She had created a program that also gave her a mathematical model of her distance so that she could take the derivative of the math function at any time point to get her “instantaneous” speed.
::你可能记得以前听说过贝卡及其轨道和田径比赛的情况。她的男朋友在她的笔记本电脑上记录了在比赛期间GPS发射机Becca所穿戴的定位信号。她正在使用一个笔记本电脑程序处理每秒一次的GPS位置信号,以便她能够确定自己在比赛期间的运行速度。这些距离和时间点是在她的笔记本电脑上绘制的。她创建了一个程序,也给了她一个距离的数学模型,以便她能够随时提取数学函数的衍生物,以获得她的“即时”速度。

What if, instead of just finding her speed at any time during the race, she wanted to find her acceleration ? How would that be done?
::万一她不想在赛跑中找到速度反而想找加速度呢?

Higher Order Derivatives
::高级命令衍生物

If the function $\begin{aligned} f \end{aligned}$ has a derivative $\begin{aligned} f^{'} \end{aligned}$ that is differentiable, then the derivative of $\begin{aligned} f^{'} \end{aligned}$ , denoted by $\begin{aligned} f^{''} \end{aligned}$ is called the second derivative of $\begin{aligned} f \end{aligned}$ . We can continue the process of differentiating and obtain third, fourth, fifth and higher derivatives of $\begin{aligned} f \end{aligned}$ . They are denoted as shown below:
::如果函数 f 的衍生物 f 不同,那么 f 的衍生物f 即 f 的第二个衍生物。我们可以继续区分并获得 f 的第三、第四、第五和更高衍生物。

**Notations for Higher Order Derivatives**
1st	2nd	3rd	4th	$\begin{aligned} n \end{aligned}$ th order
$\begin{aligned} f^{'} \end{aligned}$	$\begin{aligned} f^{''} \end{aligned}$	$\begin{aligned} f^{'''} \end{aligned}$	$\begin{aligned} f^{(4)} \end{aligned}$	%7D"> $\begin{aligned} f^{(n)} \end{aligned}$
$\begin{aligned} y^{'} \end{aligned}$	$\begin{aligned} y^{''} \end{aligned}$	$\begin{aligned} y^{'''} \end{aligned}$	$\begin{aligned} y^{(4)} \end{aligned}$	%7D"> $\begin{aligned} y^{(n)} \end{aligned}$
$\begin{aligned} \frac{d y}{d x} \end{aligned}$	$\begin{aligned} \frac{d^{2} y}{d x^{2}} \end{aligned}$	$\begin{aligned} \frac{d^{3} y}{d x^{3}} \end{aligned}$	$\begin{aligned} \frac{d^{4} y}{d x^{4}} \end{aligned}$	$\begin{aligned} \frac{d^{n} y}{d x^{n}} \end{aligned}$
$\begin{aligned} D_{x} y \end{aligned}$	$\begin{aligned} D_{x}^{2} y \end{aligned}$	$\begin{aligned} D_{x}^{3} y \end{aligned}$	$\begin{aligned} D_{x}^{4} y \end{aligned}$	$\begin{aligned} D_{x}^{n} y \end{aligned}$

Given $\begin{aligned} f (x) = - 2 x^{2} - 4 x - 1 \end{aligned}$ . What is $\begin{aligned} f^{''} (x) \end{aligned}$ ?
::给定 f ( x) = - 2 x 2 - 4 x - 1 。 F = = ( x) = = = 2 x 2 - 4 x - 1。

Recall that $\begin{aligned} f^{''} (x) \end{aligned}$ means “The second derivative of $\begin{aligned} f (x) \end{aligned}$ ”, or “The derivative of the derivative of $\begin{aligned} f (x) \end{aligned}$ ”. The function $\begin{aligned} f (x) \end{aligned}$ must be differentiated twice as follows:
::回顾f {______________________________(xx)系指“f(x)的第二个衍生物”或“f(x)的衍生物”。

\begin{aligned} f^{'} (x) & = \frac{d}{d x} (- 2 x^{2} - 4 x - 1) & \dots Determine the 1^{st} derivate . \\ = - 4 x - 4 \\ f^{''} (x) & = \frac{d}{d x} (- 4 x - 4) & \dots Determine the 2^{nd} derivate . \\ = - 4 \end{aligned}

Therefore, $\begin{aligned} f^{''} (x) = - 4 \end{aligned}$
::因此, f = = = = = 4

Examples
::实例

Example 1
::例1

Earlier, you were asked about Becca would find her acceleration. Since Becca has already created a program to calculate her instantaneous speed at a given point on the track by finding the derivative of the mathematical model to her GPS position data, she could then take the derivative of that function, the second derivative, to find her instantaneous acceleration at the same point in the race.
::早些时候,有人问你贝卡会发现她的加速度。因为贝卡已经建立了一个程序,通过将数学模型的衍生物找到GPS位置数据,计算她在轨道上某一点的瞬时速度,因此她可以选择该函数的衍生物,即第二个衍生物,在比赛的同一点找到她的瞬时加速度。

By finding her instantaneous speed and acceleration at different points in the race, she can learn a lot about her performance during the race, and hopefully target areas she needs to work on to improve her overall success.
::通过在赛事的不同点找到她的即时速度和加速度, 她可以了解她比赛期间的成绩, 希望能针对她需要努力的领域, 以提高她的总体成功率。

Example 2
::例2

Given $\begin{aligned} f (x) = (- x^{4} - 4 x^{3} - 5 x^{2} + 3) \end{aligned}$ . Find $\begin{aligned} f^{''} (x) \end{aligned}$ when $\begin{aligned} x = 3 \end{aligned}$ .
::给定 f ( x) = ( - x 4 - 4 x 3 - 5 x 2 + 3) , 当 x = 3 时查找 f ' ( x) = 3 。

Again, the function $\begin{aligned} f (x) \end{aligned}$ must be differentiated twice; then the 2 ^nd derivative must be evaluated:
::同样,函数f(x)必须两次区分;然后必须评价第2项衍生物:

\begin{aligned} f^{'} (x) & = \frac{d}{d x} (- x^{4} - 4 x^{3} - 5 x^{2} + 3) & \dots Determine the 1^{st} derivate . \\ = - 4 x^{3} - 12 x^{2} - 10 x \\ f^{''} (x) & = \frac{d}{d x} (- 4 x^{3} - 12 x^{2} - 10 x) & \dots Determine the 2^{nd} derivate . \\ = - 12 x^{2} - 24 x - 10 \\ f^{''} (3) & = - 12 (3)^{2} - 24 (3) - 10 & \dots Evaluate the 2^{nd} derivate \\ = - 190 \end{aligned}

Therefore, $\begin{aligned} f^{''} (3) = - 190 \end{aligned}$ .
::因此,f __________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Example 3
::例3

Show that $\begin{aligned} y = x^{3} + 3 x + 2 \end{aligned}$ satisfies the differential equation $\begin{aligned} y^{'''} + x y^{''} - 2 y^{'} = 0 \end{aligned}$ .
::显示 y = x 3 + 3 x + 2 满足差分方程 y = + + x y = 0。

We need to obtain the first, second, and third derivatives and substitute them into the differential equation to check for equality.
::我们需要获得第一、二和三等衍生物,并用它们来取代差异方程,以保障平等。

\begin{aligned} y & = x^{3} + 3 x + 2 \\ y^{'} & = 3 x^{2} + 3 \\ y^{''} & = 6 x \\ y^{'''} & = 6. \end{aligned}

Substituting,
::以物配主者,

\begin{aligned} y^{'''} + x y^{''} - 2 y^{'} & = 6 + x (6 x) - 2 (3 x^{2} + 3) \\ = 6 + 6 x^{2} - 6 x^{2} - 6 \\ = 0 \end{aligned}

which satisfies the equation.
::这满足了等式。

Example 4
::例4

Find the fifth derivative of $\begin{aligned} f (x) = 2 x^{4} - 3 x^{3} + 5 x^{2} - x - 1 \end{aligned}$ .
::查找 f ( x) = 2 x 4 - 3 x 3 + 5 x 2 - x 1 的第五种衍生物。

To find the fifth derivative, we must first find the first, second, third, and fourth derivatives as follows:
::为了找到第五个衍生物,我们必须首先找到第一、第二、第三和第四个衍生物如下:

$\begin{aligned} f^{'} (x) = 8 x^{3} - 9 x^{2} + 10 x - 1 \end{aligned}$
::f = (xx) = 8 x 3 - 9 x 2 + 10 x - 1

$\begin{aligned} f^{(2)} (x) = 24 x^{2} - 18 x + 10 \end{aligned}$
:f) 2) (x) = 24 x 2 - 18 x + 10

$\begin{aligned} f^{(3)} (x) = 48 x - 18 \end{aligned}$
:f) 3) (x) = 48 x - 18

$\begin{aligned} f^{(4)} (x) = 48 \end{aligned}$
:f) 4) (x) = 48

$\begin{aligned} f^{(5)} (x) = 0 \end{aligned}$
:f) 5) (x) = 0

Review
::回顾

Given: $\begin{aligned} v (x) = - 4 x^{3} + 3 x^{2} + 2 x + 3 \end{aligned}$ . What is $\begin{aligned} v^{''} (x) \end{aligned}$ ?
:x)= - 4 x 3 + 3 x 2 + 2 x + 3。

Given: $\begin{aligned} m (x) = x^{2} + 5 x \end{aligned}$ . What is $\begin{aligned} m^{''} (x) \end{aligned}$ ?
::百分比: m (x) = x 2 + 5 x. m = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Given: $\begin{aligned} d (x) = 3 x^{4} e^{x} \end{aligned}$ . What is $\begin{aligned} d^{''} (x) \end{aligned}$ ?
::d (x)= 3 x 4 e x 。 d = (x) = 3 x 4 e x 。

Given: $\begin{aligned} t (x) = - 2 x^{5} \sin (x) \end{aligned}$ . What is $\begin{aligned} \frac{d^{2} t}{d x^{2}} ? \end{aligned}$
::给出: t (x) = - 2 x 5 罪 * (x) 。 d 2 t d x 2 是什么?

Given: $\begin{aligned} y (x) = 3 x^{5} e^{x} . \end{aligned}$ What is $\begin{aligned} \frac{d^{2}}{d x^{2}} ? \end{aligned}$
::说明:y(x) = 3 x 5 e x. d 2 d x 2是什么?

Find $\begin{aligned} \frac{d^{3} y}{d x^{3}} |_{x = 1} \end{aligned}$ , where $\begin{aligned} y = \frac{2}{x^{3}} \end{aligned}$ .
::查找 d 3 y d x 3 x = 1, 其中y = 2 x 3 。

Suppose $\begin{aligned} u^{'} (0) = 98 \end{aligned}$ and $\begin{aligned} {(\frac{u}{q})}^{'} (0) = 7 \end{aligned}$ . Find $\begin{aligned} q (0) \end{aligned}$ assuming $\begin{aligned} u (0) = 0 \end{aligned}$ ?
::假设 `( 0 ) = 98 和 ( u q ) = 7。发现 q ( 0 ) = 7 。假设 u ( 0 ) = 0 = 0 ?

Given: $\begin{aligned} b (x) = \frac{x^{2} - 5 x + 4}{- 5 x + 2} \end{aligned}$ . What is: $\begin{aligned} b^{'} (2) \end{aligned}$ ?
::b (x) = x 2 = 5 x + 4 = 5 x + 5 x + 2 。

Given: $\begin{aligned} m (x) = \frac{e^{x}}{3 x + 4} \end{aligned}$ . What is $\begin{aligned} \frac{d m}{d x} \end{aligned}$ ?
::说明: m (x) = e x 3 x + 4。 d m d x 是什么?

What is $\begin{aligned} \frac{d}{d x} \cdot \frac{\sin (x)}{x - 4} \end{aligned}$ ?
::什么是 d x 罪 ? ( x) x 4 ?

Given $\begin{aligned} q (x) = \frac{x}{\sin (x)} \end{aligned}$ . What is $\begin{aligned} q^{''} (x) = \frac{x}{\sin (x)} \end{aligned}$ ?
::给定 q ( x) = x sin ( x ) 。什么是 q ( x) = x sin ( x ) ?

The position of a certain nano particle can be approximated by the function $\begin{aligned} t^{3} + t \end{aligned}$ . What function gives the acceleration of the particle?
::某个纳米粒子的位置可以被函数 t 3 + t 近似。什么函数使粒子加速?

The position of a car is given by the function $\begin{aligned} \sin (t) + 3 t^{2} \end{aligned}$ . Is the car accelerating or decelerating?
::汽车的位置是由功能罪 + 3 t 2 赋予的。汽车在加速还是减速?

The position of a velociraptor chasing a triceratops is given by the function $\begin{aligned} \cos (- t) \end{aligned}$ . Is the raptor experiencing positive or negative jerk at $\begin{aligned} t = \frac{3 π}{2} \end{aligned}$ ?
::追逐三角龙的迅猛龙的位置由函数 cos (- t) 给出。迅猛龙在 t = 3 2 时是否正或负的混蛋?

The position of the moon is night sky given by the following function of time: $\begin{aligned} \frac{1}{12} t^{4} - \frac{3}{6} t^{3} - 5 t^{2} + π^{π} \end{aligned}$ . Name a time when the moon is experiencing no acceleration at all.
::月亮的位置是夜空,由以下时间函数给定: 1 12 t 4 - 3 6 t 3 - 5 t 2 + + + + + + + 。请指出月球完全没有加速度的时间。

What is the maximum number of times one would have to differentiate a $\begin{aligned} N \end{aligned}$ -degree polynomial before the derivative becomes zero?
::在衍生物变成零之前,一个人必须区分N度多元度的最大次数是多少?

章节大纲

Higher Order Derivatives ::高级命令衍生物

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾