Course: 8.10 总和和差额的衍生物

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总和和差额的衍生物

Juan has been out playing with his model rocket all afternoon. Partway through the day, he started taking videos of the flights using his cell phone. Watching the video, he notices that the rockets actually seem to be getting faster after the launch instead of starting off at full speed and slowing down due to gravity.
::胡安整个下午都在玩他的模拟火箭。半天,他开始用他的手机拍摄飞行录像。观看录像后,他注意到火箭发射后似乎正在加快速度,而不是全速发射,并因重力而放慢速度。

Juan figures it is reasonable to assume it takes a bit for the engines to get the rocket up to full speed, but the acceleration seems to continue past when he figures that would continue.
::胡安认为引擎需要一点时间才能把火箭加速到全速, 但加速速度似乎已经过去,

After considering for a while, he wonders if the decreased mass of the rocket as it burns fuel might be the cause, assuming he knows the force generated by the engines and the starting and ending weight of the rocket, is there a way he could conjecture whether the increased acceleration might be a result of the decreased mass?
::在考虑过一段时间后,他怀疑火箭燃烧燃料时质量下降是否是原因,假设他知道发动机产生的力以及火箭的起尾重量,他是否可以猜测加速度增加是否是质量下降的结果?

Derivatives of Sums and Differences
::总和和差额的衍生物

Theorem: If $\begin{align*}f\end{align*}$ and $\begin{align*}g\end{align*}$ are two differentiable functions at $\begin{align*}x\end{align*}$ then
::定理: 如果 f 和 g 是 x x 的两种不同的函数, 那么

$\begin{align*}\frac {d}{dx}\left [{f(x)+g(x)} \right ]=\frac{d}{dx}\left [{f(x)} \right ]+\frac {d}{dx}\left [{g(x)} \right ]\end{align*}$

and
::和

$\begin{align*}\frac {d}{dx}\left [{f(x)-g(x)} \right ]=\frac{d}{dx}\left [{f(x)} \right ]- \frac {d}{dx}\left [{g(x)} \right ]\end{align*}$

In simpler notation
::更简便的符号

$\begin{align*}(f \pm g)'= f'\pm g'\end{align*}$ .

The Product Rule
::产品规则

Theorem: (The Product Rule) If $\begin{align*}f\end{align*}$ and $\begin{align*}g\end{align*}$ are differentiable at $\begin{align*}x\end{align*}$ , then
::论理产品规则)如果f和g在x时可区别,那么

$\begin{align*}\frac {d}{dx}\left [{f(x)\cdot g(x)} \right ]= f(x) \frac {d}{dx}g(x) + g(x)\frac {d}{dx}f(x)\end{align*}$

In a simpler notation
::在一个简单的符号中

$\begin{align*}(f\cdot g)'=f\cdot g'+g\cdot f'\end{align*}$

In words, The derivative of the product of two functions is equal to the first function times the derivative of the second plus the second function times the derivatives of the derivative of the first .
::换句话说,两个函数的衍生物等于第二个函数的衍生物的第一个函数乘以第二个函数的衍生物第二个函数乘以第一个函数的衍生物第二个函数乘以第一个函数的衍生物。

Keep in mind that $\begin{align*}(f\cdot g)' \neq f'+g'\end{align*}$
::记住(fäg)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Examples
::实例

Example 1
::例1

Earlier, you were asked if Juan could make a conjecture about whether increased acceleration might be a result of decreased mass.
::早些时候,你被问及胡安是否可以猜测加速度的提高可能是质量下降的结果。

Yes, he can make the conjecture. Assuming that the force is equal to the change in mass times velocity (momentum) over change in time, then using the power rule and simplifying, he can discover that the acceleration of the rocket is equal to the force minus the velocity multiplied by change in mass over time, all divided by mass, in mathematics this looks like:
::是的,他可以作出推测。假设力等于时间变化时质量时间速度(时速)的变化,然后使用动力规则并简化,他可以发现火箭加速等于力量减去速度乘以时间质量变化的倍数,所有变化均除以质量,数学中数学学的加速度看起来如下:

$\begin{align*}a = \left(\frac{F - v \left(\frac{\delta m}{\delta t}\right)}{m}\right)\end{align*}$
::a=( F- v( 立方厘米) m)

Looking at the upper-right portion of the equation, we can see that as mass decreases, the fraction $\begin{align*}\frac{\delta m}{\delta t}\end{align*}$ goes negative. Since $\begin{align*}-v\end{align*}$ is multiplied by that fraction, it goes positive, and the overall function increases, meaning the rocket accelerates.
::看着方程的上右部分,我们可以看到,当质量下降时,分数 mt 变为负数。由于-v 乘以该分数,它会呈正数,而整体功能增加,这意味着火箭加速。

Looks like Juan was right.
::看来胡安是对的

Example 2
::例2

Find the derivative: $\begin{align*}f(x)=3x^2+2x\end{align*}$ .
::查找衍生物: f(x)=3x2+2x。

Use the power rule to help:
::使用权力规则帮助:

		$\begin{align}\frac {d}{dx}\left [{3x^2+2x} \right ]\end{align}$	$\begin{align}= \frac {d}{dx}\left [{3x^2} \right ]+\frac {d}{dx}\left [{2x} \right ]\end{align}$
			$\begin{align}= 3\frac {d}{dx}\left [{x^2} \right ]+2 \frac {d}{dx}\left [{x} \right ]\end{align}$
			$\begin{align}= 3 \left [{2x} \right ]+2 \left [{1} \right ]\end{align}$
			$\begin{align}= 6x+2\end{align}$

Example 3
::例3

Find the derivative: $\begin{align*}f(x)=x^3-5x^2\end{align*}$ .
::查找衍生物: f( x) =x3 - 5x2。

Again, use the power rule to help:
::使用权力规则来帮助:

		$\begin{align}\frac {d}{dx}\left [{x^3-5x^2} \right ]\end{align}$	$\begin{align}= \frac {d}{dx}\left [{x^3} \right ]-5 \frac {d}{dx}\left [{x^2} \right ]\end{align}$
			$\begin{align}= 3x^2- 5 \left [{2x} \right ]\end{align}$
			$\begin{align}= 3x^2-10x\end{align}$

Example 4
::例4

Find $\begin{align*}\frac {dy}{dx}\end{align*}$ for $\begin{align*}y = (3x^4 + 2)(7x^3 - 1)\end{align*}$ .
::查找 y=( 3x4+2)( 7x3- 1) 的 dydx 。

There two methods to solve this problem. One is to multiply to find the product and then use the derivative of the sum rule. The second is to directly use the product rule. Either rule will produce the same answer. We begin with the sum rule.
::解决这个问题有两种方法。一种是乘以找到产品, 然后使用总和规则的衍生物。第二种是直接使用产品规则。两种规则都会产生相同的答案。我们从总和规则开始。

		y	= $\begin{align}(3x^4 + 2)(7x^3 -1)\end{align}$
			= $\begin{align}21x^7 -3x^4 + 14x^3 -2\end{align}$

Taking the derivative of the sum yields
::取总产量的衍生物

		$\begin{align}\frac {dy}{dx}\end{align}$	$\begin{align}= {147x^6-12x^3+42x^2+0}\end{align}$
			$\begin{align}= 147x^6-12x^3+42x^2\end{align}$

Now we use the product rule.
::现在我们使用产品规则。

		$\begin{align}y'\end{align}$	$\begin{align}= (3x^4+2)\cdot (7x^3-1)'+(3x^4+2)'\cdot (7x^3-1)\end{align}$
			$\begin{align}= (3x^4+2)(21x^2)+(12x^3)(7x^3-1)\end{align}$
			$\begin{align}= (63x^6+42x^2)+(84x^6-12x^3)\end{align}$
			$\begin{align}= 147x^6-12x^3+42x^2\end{align}$

Which is the same answer.
::答案是一样的

Example 5
::例5

Given: $\begin{align*} t(x) = x - 1\end{align*}$ . What is $\begin{align*}\frac{dt}{dx}\end{align*}$ when $\begin{align*}x = 0\end{align*}$ ?
::给定 : t( x) =x- 1. 当 x=0 时, dtdx 是什么 ?

By the difference rule: $\begin{align*}(x - 1)' = (x)' - (1)' = 0\end{align*}$
::区别规则x-1)________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

$\begin{align*}x' = 1\end{align*}$ ..... By the power rule
::X1. . . . . . . . . . . . . . . . . . . .

$\begin{align*}1' = 0\end{align*}$ ..... The derivative of a constant = 0
::10 .... 恒定值=0的衍生物

So when we evaluate this at x = 0, we get 1, since 1 - 0 = 1
::所以当我们用 x = 0来评估这个值时, 我们得到1, 从 1 - 0 = 1

Example 6
::例6

What is the derivative of $\begin{align*}g(x) = (-x -1)(x + 1)\end{align*}$ ?
::g(x) =(- x-1) (x+1) 的衍生物是什么?

We'll use the difference rule
::我们用差异规则

First, expand $\begin{align*}(-x -1)(x + 1) \to -x^2 -2x -1.\end{align*}$
::首先,展开(-x-1)(x+1) x2-2-x-1)。

By the difference rule: $\begin{align*}(-x^2 -2x -1)' = (-x^2)' -(2x)' -(1)' = -2x -2\end{align*}$
::依据区别规则- x2-2x- 1) (- x2) (-2x) ( 1) ( 1) ( 2x) ( 1) ( 2x) ( 2x) - 2) 。

Example 7
::例7

Given $\begin{align*} a(x) = -\pi x^{-0.54} + 6x^4\end{align*}$ . What is $\begin{align*}\frac{dy}{dx}\end{align*}$ ?
::给定 a (x) x- 0.54+6x4。什么是 dydx ?

We'll use the difference and power rules:
::我们将使用差异和权力规则:

$\begin{align*}\frac{d}{dx}(-\pi x^{-0.54} + 6x^4) =\end{align*}$
::ddx (x- 054+6x4) =

$\begin{align*}\frac{d}{dx}(-\pi x^{-0.54}) + \frac{d}{dx}(6x^4)\end{align*}$ ..... By the difference rule
::ddx( x- -0. 54) +ddx( 6x4)...。按照差数规则

$\begin{align*}\to 0.54 \pi x^{-1.54} + 24x^3\end{align*}$ ..... By the power rule
::+24x3 。根据权力规则

Example 8
::例8

What is $\begin{align*}\frac{d}{dx}[(-5x)cos (x)]?\end{align*}$
::什么是 ddx[(- 5x) cos( x)] ?

We'll use the product rule:
::我们将使用产品规则:

$\begin{align*}(pq)' = p'q + pq'\end{align*}$ .
:pq) pq+pq。

$\begin{align*}p(x) = -5x \to p'(x) = -5\end{align*}$ .... By the power rule
::p( x)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\...。根据权力规则

$\begin{align*}q(x) = cos (x) \to q'(x) = -sin (x)\end{align*}$ ..... By the power rule and simplifying
::q(x) =cos(x) {(x) }}\\\\ sin(x)...。根据权力规则和简化

So we get $\begin{align*}[(-5x)cos (x)]' = (-5)cos (x) + (-5x)[-sin (x)]\end{align*}$
::因此,我们得到[(-5x)cos(x)\\\(-5)cos(x)+(-5x)[-sin(x)]

$\begin{align*}= -5cos (x) + (5x)sin (x)\end{align*}$
::5cos(x)+(5x)sin(x)

Review
::回顾

Find the derivative using the sum/difference rule.
::使用总和/差异规则查找衍生物。

$\begin{align*}y = \frac{1} {2} (x^3 - 2x^2 + 1)\end{align*}$
::y=12(x3- 2x2+1)

$\begin{align*}y = \sqrt{2} x^3 -\frac{1} {\sqrt{2}}x^2 + 2x + \sqrt{2}\end{align*}$
::y2x3 - 1\2x2+2x%2

$\begin{align*}y = a^2 - b^2 + x^2 - a - b + x\end{align*}$ (where a , b are constants)
::y=a2-b2+x2-a-b+x(其中a、b为常数)

$\begin{align*}y = x^{-3} + \frac{1} {x^7}\end{align*}$
::y=x - 3+1x7 y=x - 3+1x7

$\begin{align*}y = \sqrt{x} + \frac{1} {\sqrt{x}} \end{align*}$
::yx+1x

$\begin{align*}f(x) = (-3x + 4)^2\end{align*}$
:xx) = (- 3x+4) 2

$\begin{align*}f(x) = -0.93x^{10} + (\pi^3x)^{\frac{-5}{12}}\end{align*}$
:xx) 0.93x10+(x3x) -512

What is $\begin{align*}\frac{d}{dx} (2x + 1)^2\end{align*}$ ?
::ddx( 2x+1) 是什么 ?

Given: $\begin{align*}a(x) = (-5x + 3)^2\end{align*}$ , what is $\begin{align*}\frac{dy}{dx}\end{align*}$ ?
::给出: a(x) = (- 5x+3) 2, 什么是 dydx ?

If $\begin{align*} v(x) = -3x^3 + 5x^2 - 2x - 3\end{align*}$ , what is $\begin{align*}v'(0)\end{align*}$ ?
::如果 v( x) 3x3+5x2-2x-3, 什么是 v_( 0) ?

Find the derivative using the product rule.
::利用产品规则寻找衍生物。

$\begin{align*}y = (x^3 - 3x^2 + x) \cdot (2x^3 + 7x^4)\end{align*}$
::y= (x3- 3x2+x) = (2x3+7x4)

$\begin{align*}y = \left (\frac{1} {x} + \frac{1} {x^2}\right ) (3x^4 - 7) \end{align*}$
::y=(1x+1x2)(3x4-7)

What is the derivative of $\begin{align*}[(-3x^2 + x + 4)(-3x - 3)]\end{align*}$ ?
::[(-3x2+x+4)(-3x-3)]的衍生物是什么?

$\begin{align*}v(x) = (3x - 3) \cdot cos (x)\end{align*}$
::v(x) = (3x-3) =cos(x)

Given: $\begin{align*}k(-2) = 0\end{align*}$ , $\begin{align*}k'(-2) = 18\end{align*}$ , find $\begin{align*}r(-2)\end{align*}$ when $\begin{align*}(kr)'(-2) = 54\end{align*}$ .
::给出: k(-2) = 0. k(-2) = 18, 在 (kr) = 54 时找到r(-2) = 54 。

Given $\begin{align*}g(x) = (4x^2 - 4x - 5)(3x - 3)\end{align*}$ , find $\begin{align*}g'(2)\end{align*}$ .
::g(x) =( 4x2- 4x- 5)( 3x- 3), 参见 g=(2)。

Find $\begin{align*}\frac{d}{dx}[(-4x + 3) \cdot sin (x)\end{align*}$ .
::查找 ddx[( (- 4x+3)) {sin( x) 。 </li> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <li> Find <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="%5Cfrac%7Bd%7D%7Bdx%7D%5B(x%5E2%20-%203)%20(-2x%5E2%20%2B%204x%20-%201)%5D"> d d x [ ( x 2 − 3 ) ( − 2 x 2 + 4 x − 1 ) ] <script id="MathJax-Element-82" type="math/tex"> \begin{align*}\frac{d}{dx}[(x^2 - 3) (-2x^2 + 4x - 1)]\end{align*} .
::查找 ddx[( (x2- 3) (-2x2+4x- 1)] 。

Given $\begin{align*}t(1) = 0\end{align*}$ , $\begin{align*}t'(1) = 17\end{align*}$ , find $\begin{align*}a(1)\end{align*}$ when $\begin{align*}(ta)'(1) = 272\end{align*}$ .
::鉴于 t(1)=0, t(1)=17,在(ta) =272时找到a(1)。

Given $\begin{align*}d(x) =(2x^2 + 3x - 1)(2x + 1)\end{align*}$ , find $\begin{align*}d'(-1)\end{align*}$ .
::给定 d( x) = ( 2x2+3x- 1)( 2x+1) , 找到 d_( 1) 。

Review (Answers)
::回顾(答复)

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。

		$\begin{align}\frac {d}{dx}\left [{3x^2+2x} \right ]\end{align}$	$\begin{align}= \frac {d}{dx}\left [{3x^2} \right ]+\frac {d}{dx}\left [{2x} \right ]\end{align}$
			$\begin{align}= 3\frac {d}{dx}\left [{x^2} \right ]+2 \frac {d}{dx}\left [{x} \right ]\end{align}$
			$\begin{align}= 3 \left [{2x} \right ]+2 \left [{1} \right ]\end{align}$
			$\begin{align}= 6x+2\end{align}$

		$\begin{align}\frac {d}{dx}\left [{x^3-5x^2} \right ]\end{align}$	$\begin{align}= \frac {d}{dx}\left [{x^3} \right ]-5 \frac {d}{dx}\left [{x^2} \right ]\end{align}$
			$\begin{align}= 3x^2- 5 \left [{2x} \right ]\end{align}$
			$\begin{align}= 3x^2-10x\end{align}$

		$\begin{align}y'\end{align}$	$\begin{align}= (3x^4+2)\cdot (7x^3-1)'+(3x^4+2)'\cdot (7x^3-1)\end{align}$
			$\begin{align}= (3x^4+2)(21x^2)+(12x^3)(7x^3-1)\end{align}$
			$\begin{align}= (63x^6+42x^2)+(84x^6-12x^3)\end{align}$
			$\begin{align}= 147x^6-12x^3+42x^2\end{align}$

Section outline

General

总和和差额的衍生物

Derivatives of Sums and Differences ::总和和差额的衍生物

The Product Rule ::产品规则

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

Example 8 ::例8

Review ::回顾

Review (Answers) ::回顾(答复)