3.5 区别规则:总和和和差异

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  区别规则:总和和差异

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  Based on your knowledge of the limit definition of the derivative of a function, and the discussed in a previous concept, can you make a prediction at this time how the derivative of a sum or difference of two functions should be determined?
  ::根据你对函数衍生物的限值定义的了解,以及前一个概念中的讨论,你目前能否预测如何确定两个函数的总和或差数的衍生物?

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  Differentiation of Sums and Differences
  ::合计和差额的差别

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  Here are the differentiation rules for the sum and difference of two functions:
  ::以下是关于两项职能之和和差的区别规则:

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  If  $\begin{array}{r}f\end{array}$ and  $\begin{array}{r}g\end{array}$ are two differentiable functions at $\begin{array}{r}x\end{array}$  then
  ::如果 f 和 g 是 x x 时两个可区别的函数, 那么

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  $$\begin{array}{rl}\frac{d}{dx}[f(x)+g(x)]& =\frac{d}{dx}[f(x)]+\frac{d}{dx}[g(x)]\\ & \text{and}\\ \frac{d}{dx}[f(x)-g(x)]& =\frac{d}{dx}[f(x)]-\frac{d}{dx}[g(x)]\end{array}$$

  
  ::ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)]和ddx[f(x)-g(x)]=dddx[f(x)]-dddx[g(x)]

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  In simpler notation
  ::更简便的符号

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  $$\begin{array}{r}(f\pm g{)}^{\mathrm{\prime }}=f^{\mathrm{\prime }}\pm g^{\mathrm{\prime }}.\end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}(fg) {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}(fg) {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}(fg) {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}(fg)

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  Using the limit properties of previous chapters should allow you to figure out why these differentiation rules apply.
  ::使用前几章的限制性能应能使你了解为什么适用这些区别规则。

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  You often need to apply multiple rules to find the derivative of a function. To find the derivative of  $\begin{array}{r}f(x)=3x^2+2x\end{array}$ , you need to apply the sum of formula and the power rule:
  ::您通常需要应用多个规则来查找函数的衍生物。要找到 f( x) =3x2+2x的衍生物,您需要应用公式和权力规则的总和:

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  $$\begin{array}{rl}\frac{d}{dx}[3x^2+2x]& =\frac{d}{dx}[3x^2]+\frac{d}{dx}[2x]\\ & =3\frac{d}{dx}[x^2]+2\frac{d}{dx}[x]\\ & =3[2x]+2[1]\\ & =6x+2\end{array}$$

  
  ::ddx[3x2+2x]=ddx[3x2]+ddx[2x]=3ddx[x2]+2ddx[x2]+2ddx[x]=3[2x]+2[1]=6x+2

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to make a prediction for the sum and differences of derivatves. 
  ::早些时候,有人要求你对衍生物的总和和差异作出预测。

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  In a previous concept, you learned that if the limits exist:
  ::在前一个概念中,你了解到,如果存在限度:

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  $$\begin{array}{r}\underset{x\to a}{lim}[f(x)\pm g(x)]=\underset{x\to a}{lim}f(x)\pm \underset{x\to a}{lim}g(x),\end{array}$$

  
  :xxxxxxxxxxxxxxxx)= limx*5xxxxxx=xxxxxxx=ag(x)x,

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  Since the derivative of a function is defined by a limit, $\begin{array}{r}\frac{d}{dx}[f(x)\pm g(x)]\end{array}$  would be defined by limit applied to $\begin{array}{r}[f(x)\pm g(x)]\end{array}$ . Work out the details to see that the above rules make sense.
  ::由于函数的衍生物由限值来定义,ddx[f(x)g(x)]将根据适用于[f(x)g(x)]的限值来定义ddx[f(x)g(x)]。

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  Example 2
  ::例2

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  Given: $\begin{array}{r}t(x)=x-1\end{array}$ , what is $\begin{array}{r}\frac{dt}{dx}\end{array}$  when $\begin{array}{r}x=0\end{array}$
  ::给定值: t( x) =x- 1, x=0 时 dtdx 是什么

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  By the difference rule: $\begin{array}{r}(x-1{)}^{\mathrm{\prime }}=(x{)}^{\mathrm{\prime }}-(1{)}^{\mathrm{\prime }}=0\end{array}$
  ::区别规则x-1)________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  $\begin{array}{r}{x}^{\mathrm{\prime }}=1\end{array}$  ..... By the power rule
  ::X1. . . . . . . . . . . . . . . . . . . .

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  $\begin{array}{r}{1}^{\mathrm{\prime }}=0\end{array}$  ..... The derivative of a constant = 0
  ::10 .... 恒定值=0的衍生物

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  So when we evaluate this at $\begin{array}{r}x=0\end{array}$ , we get 1, since $\begin{array}{r}1-0=1\end{array}$
  ::所以当我们用x=0来评估这个时, 我们从1到0=1得到1

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  Example 3
  ::例3

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

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  Use the difference and power rules to help:
  ::使用差异和权力规则帮助:

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  $$\begin{array}{rl}\frac{d}{dx}[x^3-5x^2]& =\frac{d}{dx}[x^3]-5\frac{d}{dx}[x^2]\\ & =3x^2-5[2x]\\ & =3x^2-10x\end{array}$$

  
  ::ddx[x3-5x2]=ddx[x3]-5ddx[x2]=3x2]-5[2x]=3x2-10x

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  Example 4
  ::例4

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  Given $\begin{array}{r}a(x)=-\pi x^{-0.54}+6x^4\end{array}$ . What is $\begin{array}{r}\frac{d}{dx}a(x)\end{array}$ ?
  ::给定 a( x) x- 054+6x4。 什么是 dxa( x) ?

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  We'll use the sum and power rules:
  ::我们将使用总和和权力规则:

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  $$\begin{array}{rlrl}\frac{d}{dx}(-\pi x^{-0.54}+6x^4)& =\frac{d}{dx}(-\pi x^{-0.54})+\frac{d}{dx}(6x^4)& & \dots \text{By the sum rule}\\ & =-\pi \frac{d}{dx}(x^{-0.54})+6\frac{d}{dx}(x^4)& & \dots \text{By the Constant - function Product rule}\\ & =0.54\pi x^{-1.54}+24x^3& & \dots \text{By the power rule}\end{array}$$

  
  ::ddx( x- 054+6x4) =ddx( x-0. 54) +ddx( 6x4. ). 按规则总和 ddx( x-0. 54) +6ddx( x4. ). 按常数 - 常数 - 产品规则= 0.54xx- 1.54+24x3. 根据权力规则...

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  Review
  ::回顾

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  For #1-7, find the derivative using the sum/difference rule
  ::对于 # 1-7, 使用总和/异差规则查找衍生物

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  1. $\begin{array}{r}y=\frac{1}{2}(x^3-2x^2+1)\end{array}$
     ::y=12(x3- 2x2+1)
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  2. $\begin{array}{r}y=\sqrt{2}{x}^3-\frac{1}{\sqrt{2}}{x}^2+2x+\sqrt{2}\end{array}$
     ::y=2x3- 12x2+2x+2
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  3. $\begin{array}{r}y=a^2-b^2+x^2-a-b+x\end{array}$  (where  $\begin{array}{r}a,b\end{array}$ are constants)
     ::y=a2-b2+x2-a-b+x(a,b为常数)
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  4. $\begin{array}{r}y=x^{-3}+\frac{1}{x^7}\end{array}$
     ::y=x - 3+1x7 y=x - 3+1x7
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  5. $\begin{array}{r}y=\sqrt{x}+\frac{1}{\sqrt{x}}\end{array}$
     ::y=x+1x y=x+1x
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  6. $\begin{array}{r}f(x)=(-3x+4{)}^2\end{array}$
     :xx) = (- 3x+4) 2
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  7. $\begin{array}{r}f(x)=-0.93x^{10}+({\pi }^{3}x{)}^{\frac{-5}{12}}\end{array}$
     :xx) 0.93x10+(x3x) -512
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  8. What is $\begin{array}{r}\frac{d}{dx}(2x+1{)}^2\end{array}$ ?
     ::ddx( 2x+1) 是什么 ?
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  9. Given:  $\begin{array}{r}a(x)=(-5x+3{)}^2\end{array}$ What is $\begin{array}{r}\frac{dy}{dx}\end{array}$ ?
     ::说明:a(x)=(-5x+3)2
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  10. $\begin{array}{r}v(x)=-3x^3+5x^2-2x-3\end{array}$  What is $\begin{array}{r}{v}^{\mathrm{\prime }}(0)\end{array}$ ?
      ::v(x) 3x3+5x2-2x-3 什么是 v_(0)?
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  11. $\begin{array}{r}f(x)=2x^2+3x+1\end{array}$ . Find $\begin{array}{r}{f}^{\mathrm{\prime }}(x)\end{array}$ .
      ::f(x) = 2x2+3x+1. 查找 f_(x) 。
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  12. $\begin{array}{r}f(x)=\frac{1}{\sqrt{x}}-\frac{1}{x}\end{array}$ . Find $\begin{array}{r}{f}^{\mathrm{\prime }}(1)\end{array}$ .
      ::f(x) = 1x- 1x. 查找 f_(1) 。
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  13. $\begin{array}{r}y=(x+1)(x+2)\end{array}$ . Evaluate  $\begin{array}{r}\frac{dy}{dx}\end{array}$ at $\begin{array}{r}x=-\frac{1}{2}\end{array}$
      ::y= (x+1) (x+2) 。 x @ 12 时评价 dydx 值 。
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  14. $\begin{array}{r}f(x)=2a{x}^3+x^2;f^{\mathrm{\prime }}(-2)=0\end{array}$ . Find $\begin{array}{r}a\end{array}$ .
      ::f( x) = 2ax3+x2; f}( 2) = 0。 查找 a 。
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  15. $\begin{array}{r}f(x)=a(x^2-5)\end{array}$ ; find  $\begin{array}{r}a\end{array}$ so that $\begin{array}{r}{f}^{\mathrm{\prime }}(5)=20\end{array}$ .
      ::f(x) =a(x2-5); 找到f(5) =20。

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  Review (Answers)
  ::回顾(答复)

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