引引法规则和高级衍生物

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You may recall hearing about Becca and her Track and Field competition in a prior lesson. Her boyfriend had taken a picture of her just as she started to pull away from the others on the track. We learned how she might learn to identify her instantaneous speed at just the split second the picture was taken by using calculus to find a derivative.
::你可能记得在上一堂课中听说过贝卡和她的田径和田径比赛。她的男朋友在她开始离开赛道上其他人时就拍了一张她的照片。我们学到了她如何学会如何在照片拍摄的瞬间分秒里通过微积分找到衍生物。

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What if, instead of just finding her speed at that split second, she wanted to find her acceleration ?
::万一她没有在瞬间找到速度 反而想找加速度呢?

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Quotient Rule and Higher Derivatives
::引引法规则和高级衍生物

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The Quotient Rule
::引引规则

Theorem: (The Quotient Rule) If f and g are differentiable functions at x and g ( x ) ≠ 0, then $\begin{align*}\frac {d}{dx}\left [ \frac{f(x)}{g(x)} \right ]= \frac {g(x) \frac {d}{dx}\left [{f(x)} \right ] - f(x) \frac{d}{dx} \left [{g(x)} \right ]}{\left [{g(x)} \right ]^2}\end{align*}$ 播放段落 In simpler notation ::更简便的符号 $\begin{align*}\left ( \frac{f}{g} \right )'=\frac {g\cdot f'-f\cdot g'}{g^2}\end{align*}$ 播放段落 Keep in mind that the order of operations is important (because of the minus sign in the numerator) and $\begin{align*}\left ( \frac{f}{g} \right )'\neq \frac {f'}{g'}\end{align*}$ . ::切记操作的顺序很重要(因为分子中的减号)和(fg)\\f\\\g\\\\\f\h\h\h\h\h\h\h\h\h\h\h\h\h\h\h\h\h。

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Higher Derivatives
::高衍生物

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If the derivative $\begin{align*}\,\! f'\end{align*}$ of the function $\begin{align*}\,\! f\end{align*}$ is differentiable, then the derivative of $\begin{align*}f'\end{align*}$ , denoted by $\begin{align*}\,\! f''\end{align*}$ is called the second derivative of $\begin{align*}\,\! f\end{align*}$ . We can continue the process of differentiating and obtain third, fourth, fifth and higher derivatives of $\begin{align*}\,\! f\end{align*}$ . They are denoted by $\begin{align*}\,\! f'\end{align*}$ , $\begin{align*}\,\! f''\end{align*}$ , $\begin{align*}\,\! f'''\end{align*}$ , $\begin{align*}\,\! f^{(4)}\end{align*}$ , $\begin{align*}\,\! f^{(5)}\end{align*}$ , . . . ,
::如果函数 f 的衍生物 f 的衍生物是不同的,那么 f 的衍生物就是 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、第五及更高衍生物。 这些衍生物由 f f 、 f 、 f 、 f 、 f 、 f 、 f 、 f 、 f 、 f 、 f .

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked how Becca could find her acceleration in addition to her speed.
::之前有人问过你 Becca除了速度之外 如何能发现加速度

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Once Becca has calculated her instantaneous speed at a given point on the track by finding the derivative, she could then take the derivative of that function to find her instantaneous acceleration at the same point in the race.
::一旦贝卡通过找到衍生物在轨道上某一点计算出其瞬时速度后,她就可以利用该函数的衍生物在比赛的同一点发现其瞬时加速度。

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By finding her instantaneous speed and acceleration at different points in the race, she can learn a lot about what points made a difference in her overall success, and also what points she needs to work on.
::通过在比赛的不同点上发现她的即时速度和加速度,她可以了解很多因素,了解哪些因素改变了她的总体成功,以及哪些因素她需要努力。

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Example 2
::例2

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Find $\begin{align*}= \frac {dy}{dx}\end{align*}$ for $\begin{align*}y = \frac {x^2-5}{x^3+2}\end{align*}$ .
::查找 y=x2 - 5x3+2 的 = dydx。

	$\begin{align*}\frac {dy}{dx}\end{align*}$	$\begin{align*}= \frac {d}{dx}\left [ \frac{x^2-5}{x^3+2} \right ]\end{align*}$
		$\begin{align*}= \frac {(x^3+2).(x^2-5)'-(x^2-5).(x^3+2)'}{(x^3+2)^2}\end{align*}$
		$\begin{align*}= \frac {(x^3+2)(2x)-(x^2-5)(3x^2)}{(x^3+2)^2}\end{align*}$
		$\begin{align*}= \frac {2x^4+4x-3x^4+15x^2}{(x^3+2)^2}\end{align*}$
		$\begin{align*}= \frac {-x^4+15x^2+4x}{(x^3+2)^2}\end{align*}$
		$\begin{align*}= \frac {x(-x^3+15x+4)}{(x^3+2)^2}\end{align*}$

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Example 3
::例3

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At which point(s) does the graph of $\begin{align*}y = \frac {x} {x^2+9}\end{align*}$ have a horizontal tangent line?
::y=xx2+9 的图形在哪个点有水平正切线 ?

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Since the slope of a horizontal line is zero, and since the derivative of a function signifies the slope of the tangent line, then taking the derivative and equating it to zero will enable us to find the points at which the slope of the tangent line is equal to zero, i.e., the locations of the horizontal tangents. Notice that we will need to use the quotient rule here:
::由于水平线的斜坡为零,并且由于函数的衍生物表示正切线的斜坡,然后将衍生物乘以正切线,将其等同为零,将使我们能够找到正切线的斜坡等于零的点,即横向正切点的位置。请注意,我们需要在此使用商数规则:

	$\begin{align*}y\end{align*}$	$\begin{align*}= \frac{x} {x^2 + 9}\end{align*}$
	$\begin{align*}y'\end{align*}$	$\begin{align*}= \frac{(x^2 + 9) \cdot f'(x) - x \cdot g' (x^2 + 9)} {(x^2 + 9)^2} = 0\end{align*}$	$\begin{align*}= \frac{(x^2 + 9) (1) - x (2x)} {(x^2 + 9)^2} = 0\end{align*}$

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Multiply both sides by $\begin{align*}(x^2 +9)^2\end{align*}$ ,
::用(x2+9) 乘以(x2+9) 2, 乘以两边, 乘以(x2+9) 2 ;

	$\begin{align*}x^2 + 9 - 2x^2\end{align*}$	$\begin{align*}= 0\end{align*}$
	$\begin{align*}x^2\end{align*}$	$\begin{align*}= 9\end{align*}$
	$\begin{align*}x\end{align*}$	$\begin{align*}= \pm 3\end{align*}$

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Therefore, at $\begin{align*}x = -3\end{align*}$ and $\begin{align*}x = 3\end{align*}$ , the tangent line is horizontal.
::因此,Ax*%3和x=3,正切线是水平的。

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Example 4
::例4

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Find the fifth derivative of $\begin{align*}f(x) = 2x^4 - 3x^3 + 5x^2 - x - 1\end{align*}$ .
::查找 f( x) = 2x4 - 3x3+5x2 - x- 1 的第五个衍生物。

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To find the fifth derivative, we must first find the first, second, third, and fourth derivatives.
::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。

	$\begin{align*}f'(x)\end{align*}$	= $\begin{align*}8x^3 - 9x^2 + 5x - x\end{align*}$
	$\begin{align*}f'' (x)\end{align*}$	= $\begin{align*}24x^2 - 18x + 5\end{align*}$
	$\begin{align*}f''' (x)\end{align*}$	= $\begin{align*}48x - 18\end{align*}$
	$\begin{align*}f^{(4)} (x)\end{align*}$	= $\begin{align*}48\end{align*}$
	$\begin{align*}f^{(5)} (x)\end{align*}$	=

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Example 5
::例5

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Suppose y'(2) = 0 and (y/q)(2) = 0. Find q(2), assuming y(2) = 0.
::假设 Y'(2) = 0 和 (y/q) (2) = 0。 查找 q(2) , 假设 y(2) = 0 。

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Begin with the quotient rule:
::从商数规则开始 :

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$\begin{align*}\left(\frac{y}{q}\right)' (2) = \left(\frac{y'(2)q(2) - y(2)q'(2)}{q(2)^2}\right)\end{align*}$ ..... Substitute
:yq) (2)=(yä(2)q(2)-y(2)q(2)(2)).替代物

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$\begin{align*}(0) = \left(\frac{(0)q(2) - (0)q'(2)}{q(2)^2}\right)\end{align*}$ ..... Substituting again with given values
:0) = (0) = (0) q(2)- (0) q§(2) q(2)2)...。重置给定值

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$\begin{align*}0 = \left(\frac{(0)q(2)}{q(2)^2}\right)\end{align*}$ ..... Simplify with: $\begin{align*}(0)q'(2) = 0\end{align*}$
::0=(0)q(2)q(2)2)...。 简化为: (0)q(2)=0

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$\begin{align*}0 = \frac{0}{q(2)}\end{align*}$
::0=0q(2)

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$\begin{align*}q(2) = 0\end{align*}$
::q(2)=0

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Example 6
::例6

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Find the derivative of $\begin{align*}k(x) = \frac{-2x - 4}{e^x}\end{align*}$ .
::查找 k( x)\\% 2x- 4ex 的衍生物 。

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Use the quotient rule: Note: $\begin{align*}(-2x - 4)' = -2\end{align*}$ and $\begin{align*}(e^x)' = e^x\end{align*}$
::使用商数规则: 注-2x- 4)\\\\\\\\\\\\\\\\\\\\(ex)\\\ ex

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$\begin{align*}\left(\frac{-2x-4}{e^x}\right)' = \frac{(-2)(e^x) - (-2x - 4)(e^x)}{e^{2x}}\end{align*}$ ..... Substitute
:-2x-4ex)(-2)(ex)-(-2x-4)(ex)-(-2x-4(ex)e2x.)

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$\begin{align*}\frac{2x + 2}{e^x}\end{align*}$ ..... Simplify
::2x+2ex.... 简化

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Example 7
::例7

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Given $\begin{align*}f(x) = (-x^4 -4x^3 - 5x^2 +3)\end{align*}$ . Find $\begin{align*}f''(x)\end{align*}$ when $\begin{align*}x=3\end{align*}$ .
::给定 f( x) = (- x4 - 4x3 - 5x2+3) 查找 f (x) 当 x= 3 时查找 f (x) 。

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Recall that $\begin{align*}f''(x)\end{align*}$ means "The derivative of the derivative of x "
::回顾f(x)意指"x的衍生物的衍生物"

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$\begin{align*}f'(x) = -4x^3 -12x^2 - 10x\end{align*}$ ..... Use the power rule on f(x)
::f\ (x)\\\\\\\\\ _ 4x3 - 12x2 - 10x.... 。 使用 f( x) 的权力规则

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$\begin{align*}f''(x) = -12x^2 - 24x - 10\end{align*}$ ..... Use the power rule on f'(x)
::\\\\\\\\\\\\\\\\24x- 10\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ f'\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\F\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\在\\\F\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\权力规则的权力规则的规则, 规则,f\\\\\xx

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$\begin{align*}f''(3) = -12(3)^2 -24(3) - 10 \to -108 -72 - 10 = -190\end{align*}$ ..... Substitute 3
::f(3)12(3)2-24(3)-10108-72-10190。

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$\begin{align*}\therefore f''(3) = -190\end{align*}$
::=============================================================================================================================== ==========================================================================================================================

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Review
::回顾

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Use the quotient rule to solve:
::使用商数规则解析 :

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1. Suppose $\begin{align*}u'(0) = 98\end{align*}$ and $\begin{align*}(\frac{u}{q})'(0) = 7\end{align*}$ . Find $\begin{align*}q(0)\end{align*}$ assuming $\begin{align*}u(0) = 0\end{align*}$ .
   ::假设 u_(0)=98和(uq)_(0)=7. 假设 u(0)=0,查找 q(0) 。
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2. Given: $\begin{align*} b(x) = \frac{x^2 - 5x + 4}{-5x + 2}\end{align*}$ , what is: $\begin{align*} b'(2)\end{align*}$ ?
   ::b(x)=x2-5x+4-5x+2,什么是b=(2)?
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3. Given: $\begin{align*}m(x) = \frac {e^x}{3x + 4}\end{align*}$ , what is $\begin{align*} \frac{dm}{dx}\end{align*}$ ?
   ::给定 : m( x) =ex3x+4, 什么是 dmdx ?
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4. What is $\begin{align*}\frac{d}{dx} \cdot \frac{sin (x)}{x - 4}\end{align*}$ ?
   ::ddxsin(x)x-4是什么?
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5. Find the derivative of $\begin{align*}q(x) = \frac{x}{sin (x)}\end{align*}$ .
   ::查找 q( x) =xsin( x) 的衍生物。

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Solve these higher order derivatives:
::解决这些更高订单衍生物:

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6. Given: $\begin{align*}v(x) = -4x^3 + 3x^2 + 2x + 3\end{align*}$ , what is $\begin{align*}v''(x)\end{align*}$ ?
   ::给定 : v( x) 4x3+3x2+2x+3, 什么是 v{( x) ? </span> </li> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <li> Given: <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="m(x)%20%3D%20x%5E2%20%2B%205x%20"> <span class="MathJax_Preview" style="color: inherit;"> <span class="MJXp-math" id="MJXp-Span-1347"> <span class="MJXp-mtable" id="MJXp-Span-1348"> <span> <span class="MJXp-mtr" id="MJXp-Span-1349" style="vertical-align: baseline;"> <span class="MJXp-mtd" id="MJXp-Span-1350" style="text-align: right;"> <span class="MJXp-mi MJXp-italic" id="MJXp-Span-1351"> m </span> <span class="MJXp-mo" id="MJXp-Span-1352" style="margin-left: 0em; margin-right: 0em;"> ( </span> <span class="MJXp-mi MJXp-italic" id="MJXp-Span-1353"> x </span> <span class="MJXp-mo" id="MJXp-Span-1354" style="margin-left: 0em; margin-right: 0em;"> ) </span> <span class="MJXp-mo" id="MJXp-Span-1355" style="margin-left: 0.333em; margin-right: 0.333em;"> = </span> <span class="MJXp-msubsup" id="MJXp-Span-1356"> <span class="MJXp-mi MJXp-italic" id="MJXp-Span-1357" style="margin-right: 0.05em;"> x </span> <span class="MJXp-mn MJXp-script" id="MJXp-Span-1358" style="vertical-align: 0.5em;"> 2 </span> </span> <span class="MJXp-mo" id="MJXp-Span-1359" style="margin-left: 0.267em; margin-right: 0.267em;"> + </span> <span class="MJXp-mn" id="MJXp-Span-1360"> 5 </span> <span class="MJXp-mi MJXp-italic" id="MJXp-Span-1361"> x </span> </span> </span> </span> </span> </span> </span> <span class="MathJax_SVG MathJax_SVG_Processing" id="MathJax-Element-80-Frame" style="font-size: 100%; display: inline-block;" tabindex="-1"> </span> <script id="MathJax-Element-80" type="math/tex"> \begin{align*}m(x) = x^2 + 5x \end{align*} , what is $\begin{align*}m''(x)\end{align*}$ ?
   ::给定 : m( x) =x2+5x, 什么是 m( x) ?
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7. Given: $\begin{align*}d(x) = 3x^4e^x\end{align*}$ , what is $\begin{align*}d''(x)\end{align*}$ ?
   ::给出: d( x) = 3x4ex, 什么是 d( x) ?
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8. Given: $\begin{align*}t(x) = -2x^5sin (x)\end{align*}$ , what is $\begin{align*}\frac{d^2t}{dx^2}\end{align*}$ ?
   ::给定 : t( x) % 2x5sin( x), 什么是 d2tdx2 ?
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9. What is $\begin{align*}\frac{d^2} {dx^2}3x^5e^x\end{align*}$ ?
   ::D2dx23x5ex 是什么?

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Solve:
::解决 :

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11. Find the derivative of $\begin{align*}y = \frac{3} {\sqrt{x} + 3}\end{align*}$ .
    ::查找 y= 3x+3 的衍生物。
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12. Find the derivative of $\begin{align*}y = \frac{4x + 1} {x^2 - 9}\end{align*}$ .
    ::查找 y= 4x+1x2- 9 的衍生物。
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13. Newton’s Law of Universal Gravitation states that the gravitational force between two masses (say, the earth and the moon), m and M is equal two their product divided by the squared of the distance r between them. Mathematically, $\begin{align*}F = G\frac{mM} {r^2}\end{align*}$ where G is the Universal Gravitational Constant (1.602 × 10 ^-11 Nm ² /kg ² ). If the distance r between the two masses is changing, find a formula for the instantaneous rate of change of F with respect to the separation distance r .
    ::牛顿的《万国引力法》规定,两个质量(如地球和月球)之间, m和M和M之间的引力等于两个质量(如地球和月球)之间的引力,它们的产品除以它们之间的距离平方。从数学角度讲,F=GmMMM2,其中G是普遍引力常数(1.602 × 10-11Nm2/kg2)。如果两个质量之间的距离正在变化,则在分离距离r方面找到F瞬时变化速度的公式。
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14. Find $\begin{align*}\frac{d} {d \psi} \left [\frac{\psi \psi_0 + \psi^3} {3 - \psi_0}\right ]\end{align*}$ , where $\begin{align*}\psi_0\end{align*}$ is a constant.
    ::查找 dd[030], 其中 0 是常数 。
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15. Find $\begin{align*}\frac{d^3y} {dx^3} |_{x = 1},\end{align*}$ where $\begin{align*}y = \frac{2} {x^3}\end{align*}$ .
    ::查找 y= 2x3 所在的 d3ydx3\\ x= 1 。

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Review (Answers)
::回顾(答复)

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Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。