3.7 三角函数的衍生物

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  三角函数的衍生物

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  The functions $\begin{array}{r}\sin x\end{array}$  and $\begin{array}{r}\cos x\end{array}$  are periodic, with period $\begin{array}{r}2\pi \end{array}$ . You have learned that the derivative of a differentiable function gives the slope of the tangent line at a point. Before proceeding with this lesson, look at the function curves for the two functions, and see if you can identify any points where you know what the derivative will be. For each function, do these sets of points repeat as $\begin{array}{r}x\end{array}$  increases or decreases? How often? Can you make a general statement about the of the trigonometric functions?
  ::函数 sinx 和 cosx 是周期性的, 时段为 2 。 您已经知道, 一个不同函数的衍生物会给一个点的正切线的斜度。 在继续这个课程之前, 请先查看这两个函数的函数曲线, 并查看您是否能够确定哪个点知道该函数会是什么。 对于每个函数, 这些组合的点会重复 x 增加或减少吗? 多久? 您能够对三角函数做一般性说明吗 ?

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  Derivatives of Trigonometric Functions
  ::三角函数的衍生物

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  We now want to find an expression for the derivative of each of the six trigonometric functions:
  ::我们现在想找到六种三角函数中每种函数的衍生物的表达方式:

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  $$\begin{array}{rl}& \sin x\cos x\tan x\\ & \csc x\sec x\cot x\end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不

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  We first consider the problem of differentiating $\begin{array}{r}\sin x\end{array}$ , using the definition of the derivative.
  ::我们首先考虑使用衍生物定义来区分sinx的问题。

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  $$\begin{array}{r}\frac{d}{dx}[\sin x]=\underset{h\to 0}{lim}\frac{\sin (x+h)-\sin x}{h}\end{array}$$

  
  ::ddx[sinx] =limh0sin}(x+h) -sinxh =limh0sin}(x+h) -sinxh

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  Since
  ::自

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  $$\begin{array}{r}\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta .\end{array}$$

  
  :)=sincoscoscossin。

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  The derivative becomes
  ::衍生物变成

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  $$\begin{array}{rl}\frac{d}{dx}[\sin x]& =\underset{h\to 0}{lim}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ & =\underset{h\to 0}{lim}\left[\sin x\left(\frac{\cos h-1}{h}\right)+\cos x\left(\frac{\sin h}{h}\right)\right]\\ & =-\sin x\cdot \underset{h\to 0}{lim}\left(\frac{1-\cos h}{h}\right)+\cos x\cdot \underset{h\to 0}{lim}\left(\frac{\sin h}{h}\right)\\ & =-\sin x\cdot (0)+\cos x\cdot (1)\\ & =\cos x.\end{array}$$

  
  ::ddx[sinx] =limh_0sin_xcosin_xin_h_xh_xh=limh#0 [sin_xxxxx(cosh-1h) +cosíxxxx(sinhhh) +cos_xxxilimh_0(sin_hh) +sin_xx_(0) +cos_xx__(1)=cosxx。

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  Let’s look at  %7D%7Bh%7D"> $\begin{array}{r}\underset{h\to 0}{lim}\frac{1-\cos (h)}{h}\end{array}$ and %7D%7Bh%7D"> $\begin{array}{r}\underset{x\to 0}{lim}\frac{\sin (h)}{h}\end{array}$ . While there are analytical methods that can be used to evaluate these two limits, let’s look at function graphs and table data. The graphs of the two functions and some calculator table data are shown below. Inspection of these near  $\begin{array}{r}x=0\end{array}$ appears to show that the limits are:
  ::让我们看看 limh01-cosh 和 limx0sinh。 虽然有一些分析方法可以用来评估这两个限制, 但让我们看看函数图表和表格数据。 下面显示两个函数的图表和一些计算器表格数据。 对这些接近 x=0的图表的检查似乎显示这些限制是:

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  -1%7D%7Bh%7D%3D0"> $\begin{array}{r}\underset{x\to 0}{lim}\frac{\cos (h)-1}{h}=0\end{array}$ and %7D%7Bh%7D%3D1"> $\begin{array}{r}\underset{h\to 0}{lim}\frac{\sin (h)}{h}=1\end{array}$
  :h) -1小时=0 和 limh=0sin=1 h=1

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  The above results can be confirmed analytically .
  ::上述结果可在分析上得到证实。

  播放段落 $\begin{array}{r}x(rad)\end{array}$ ::x( rad)	-0.001	-0.0001	0	0.0001	0.0001
  播放段落 $\begin{array}{r}f(x)\end{array}$ :fx)	-0.0005	-0.00005	播放段落 Error ::错误错误错误	0.00005	0.0005
  播放段落 $\begin{array}{r}g(x)\end{array}$ :g) (x)	0.99999833	0.999999998	播放段落 Error ::错误错误错误	0.999999998	0.99999833

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  Therefore,
  ::因此,

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  $\begin{array}{r}\frac{d}{dx}[\sin x]=\cos x\end{array}$ .
  ::ddx[sinx]=cosx。

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  It will be left as an exercise to prove that $\begin{array}{r}\frac{d}{dx}[\cos x]=-\sin x\end{array}$ .
  ::这将留作一项练习,以证明 ddx[cosx] {sinx} 。

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  The derivatives of the  six trigonometric functions are shown below.
  ::六个三角函数的衍生物如下所示。

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  $$\begin{array}{rlrlrl}& \frac{d}{dx}[\sin x]=\cos x& & \frac{d}{dx}[\cos x]=-\sin x& & \frac{d}{dx}[\tan x]={\sec }^{2}x\\ & \frac{d}{dx}[\csc x]=-\csc x\cot x& & \frac{d}{dx}[\sec x]=\sec x\tan x& & \frac{d}{dx}[\cot x]=-{\csc }^{2}x\end{array}$$

  
  ::ddx[sinx] = cos*xdx[cos*x] = cos*xdx[cos*x] [cos*xdx] =sec*x*x*xdx[sec*x*x] = sec*x*x*x*xdx[ct*x] *csc2}xdx*xx*xx*=sec*x*xxx*xdx[ct*x]

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  Keep in mind that the argument  $\begin{array}{r}x\end{array}$ for all the trigonometric functions is measured in radians.
  ::铭记所有三角函数的参数 x 用弧度测量。

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  All of the derivatives can be proved by the definition of the derivative, but the reciprocal functions can be found using a simpler method. The proof of $\begin{array}{r}\frac{d}{dx}[\tan x]={\sec }^{2}x\end{array}$    is as follows:
  ::衍生物的定义可以证明所有衍生物,但使用更简单的方法可以找到对等功能。 ddx[tanx]=sec2x的证明如下:

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  Since
  ::自

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  $$\begin{array}{r}\tan x=\frac{\sin x}{\cos x},\end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不!

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  then
  ::时当时

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  $$\begin{array}{r}\frac{d}{dx}[\tan x]=\frac{d}{dx}\left[\frac{\sin x}{\cos x}\right]\end{array}$$

  
  ::ddx [tanx] =ddx [sinxcosx]

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  Using the quotient rule ,
  ::使用商数规则,

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  $$\begin{array}{rl}& =\frac{(\cos x)(\cos x)-(\sin x)(-\sin x)}{{\cos }^{2}x}\\ & =\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}\\ & =\frac{1}{{\cos }^{2}x}\\ & ={\sec }^{2}x\end{array}$$

  
  ::= (cosx) (cosx) -(sinx) (-sinx) cos2 x=cos2x+sin2xxxcos2}x=1cos2xx=sec2}x=xsec2}xx=xxxxxxxxxxxxxxxxxc2xxxxxxxxxx

     

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked if there are any repeating points for the derivatives of trigonometric functions and if so, how often they repeat.
  ::早些时候,有人问到,三角函数衍生物是否有重复点,如果有,重复次数如何。

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  First, see if you can identify any points where you know the derivative of $\begin{array}{r}\sin x\end{array}$ and $\begin{array}{r}\cos x\end{array}$ : Each function has two places in the interval  $\begin{array}{r}0\le x\le 2\pi \end{array}$ where the tangent line has a slope of 0.
  ::首先,请看您能否辨别已知 sinx 和 osx 的衍生物的任何点 : 每个函数在间距 0 x2中有两个位置, 其中正切线的斜度为 0 。

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  For each function, do these sets of points repeat as  $\begin{array}{r}x\end{array}$ increases or decreases? Yes, these 0 slope points do repeat for each function.
  ::对于每个函数,这些数组点是重复 x 增加还是减少?是的,这些0个斜点是重复每个函数的。

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  How often? The pair of 0 slope points repeats with a period of $\begin{array}{r}2\pi \end{array}$ .
  ::10个斜点的两对重复了2的周期。

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  Can you make a general statement about the derivatives of the trigonometric functions? Because the function values repeat every period, the derivative of each function at a specific point should also repeat with period $\begin{array}{r}2\pi \end{array}$ .
  ::您能否对三角函数的衍生物做一个一般性说明?因为函数值会重复每个周期,因此每个函数在特定点的衍生物也应重复到 2+ 周期。

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  Example 2
  ::例2

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  Find $\begin{array}{r}{f}^{\mathrm{\prime }}(x)\end{array}$ if $\begin{array}{r}f(x)=x^2\cos x+\sin x\end{array}$ .
  ::如果 f( x) =x2cos *x+sin*x, 请查找 f_ (x) 。

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  Using the product rule and the formulas above, we obtain
  ::使用上述产品规则和公式,我们获得

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  $$\begin{array}{rl}{f}^{\mathrm{\prime }}(x)& =x^2(-\sin x)+2x\cos x+\cos x\\ & =-x^2\sin x+2x\cos x+\cos x.\end{array}$$

  
  ::f*(x) =x2(- 辛x) +2x科斯x(x) +2x科斯x(x) +科斯x(x) x(x) +2x科斯x(x) x) +2xxx(x) x(x) +xxx(x) x(x) =x(x) =x(x) 2(- 辛x) +2xx科斯x(x) +科斯*x(x) x(x)x+2xx(x) x) x(x) +xxxxxxx(x) x(x) +cos(x)x(x) x(x) =x(x) x) =x(x) 。

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  Example 3
  ::例3

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  Find  $\begin{array}{r}\frac{dy}{dx}\end{array}$ if $\begin{array}{r}y=\frac{\cos x}{1-\tan x}\end{array}$ . What is the slope of the tangent line at $\begin{array}{r}x=\frac{\pi }{3}\end{array}$ ?
  ::如果 y= cosx1 - tanx, 则查找 dydx。 x\\ 3 点的正切线的斜度是什么 ?

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  Using the quotient rule and the formulas above, we obtain
  ::使用商数法规则和上述公式,我们获得

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  $$\begin{array}{rl}\frac{dy}{dx}& =\frac{(1-\tan x)(-\sin x)-(\cos x)(-{\sec }^{2}x)}{(1-\tan x{)}^2}\\ & =\frac{-\sin x+\tan x\sin x+\cos x{\sec }^{2}x}{(1-\tan x{)}^2}\end{array}$$

  
  ::dydx= (1 - tanx) (- 辛x) - (cosx) (- sec2x) (1 - tanx) 2 sin*x +tanxin* x+ cosxsec2 x(1 - tanx) 2

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  To calculate the slope of the tangent line, we simply substitute $\begin{array}{r}x=\frac{\pi }{3}\end{array}$ :
  ::为了计算正切线的斜坡, 我们只需替换 x% 3 :

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  $$\begin{array}{r}\frac{dy}{dx}{|}_{x=\frac{\pi }{3}}=\frac{-\sin \left(\frac{\pi }{3}\right)+\tan \left(\frac{\pi }{3}\right)\sin \left(\frac{\pi }{3}\right)+\cos \left(\frac{\pi }{3}\right){\sec }^2\left(\frac{\pi }{3}\right)}{{\left(1-\tan \left(\frac{\pi }{3}\right)\right)}^2}.\end{array}$$

  We finally get the slope to be approximately
  ::

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  $$\begin{array}{r}\frac{dy}{dx}{|}_{x=\frac{\pi }{3}}=\mathrm{4.9.}\end{array}$$

  
  ::迪德克斯xxxxx3=4.9。

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  Example 4
  ::例4

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  Find  $\begin{array}{r}\frac{dy}{dx}\end{array}$ if $\begin{array}{r}y=\frac{\cot x}{\sin x}\end{array}$ .
  ::如果y=cot=xsin=xx,请查找 dydx 。

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  $$\begin{array}{rlrl}\frac{d}{dx}\left[\frac{\cot x}{\sin x}\right]& =\frac{\sin x\frac{dy}{dx}[\cot x]-\cot x\frac{dy}{dx}[\sin x]}{{\sin }^{2}x}& & \text{{....}Quotient Rule}\\ & =\frac{\sin x\cdot [-{\csc }^{2}x]-\cot x\cdot [\cos x]}{{\sin }^{2}x}& & \text{{.....}Trigonometric derivatives}\\ & =\frac{-\sin x\cdot [{\csc }^{2}x+{\cot }^{2}x]}{{\sin }^{2}x}& & \text{{.....}Factor}-\sin x\\ & =-\csc x\cdot [1+{\cot }^{2}x+{\cot }^{2}x]& & \text{{....}Simplify and use trig identities}\\ & =-\csc x\cdot [1+2{\cot }^{2}x]\end{array}$$

  
  ::ddx[cotxinxxxx] =sinxx[ctxx] -ctxdx[sin_xxx]2x{.\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ x\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Therefore, $\begin{array}{r}\frac{dy}{dx}=-\csc x\cdot [1+2{\cot }^{2}x]\end{array}$ .
  ::因此,didxcscx[1+2cot2x]。

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  Review
  ::回顾

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  For #1-10, find the derivative  $\begin{array}{r}{y}^{\mathrm{\prime }}\end{array}$ .
  ::10号,找到衍生物

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  1. $\begin{array}{r}y=x\sin x+2\end{array}$
     ::y=xsinx+2
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  2. $\begin{array}{r}y=x^2\cos x-x\tan x-1\end{array}$
     ::y=x2cosx-xtanx-1
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  3. $\begin{array}{r}y={\sin }^{2}x\end{array}$
     ::y=in2x
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  4. $\begin{array}{r}y=\frac{\sin x-1}{\sin x+1}\end{array}$
     ::y=sin%x - 1sin%x+1 y=sin%x - 1sin%x+1
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  5. $\begin{array}{r}y=\frac{\cos x+\sin x}{\cos x-\sin x}\end{array}$
     ::y=cos*x+sin*xcos*xx*sin*x
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  6. $\begin{array}{r}y=\frac{\sqrt{x}}{\tan x}+2\end{array}$
     ::y=xtanx+2
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  7. $\begin{array}{r}y=\csc x\sin x+x\end{array}$
     ::y= yc=xsin=xx+x+x y=csc=xsin=xxx+x
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  8. $\begin{array}{r}y=\frac{\sec x}{\csc x}\end{array}$
     ::y= y=sec=xcsc=xx y= y=sec=xc=xcsc=xx
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  9. If $\begin{array}{r}y=\csc x\end{array}$ , find $\begin{array}{r}{y}^{\mathrm{\prime }}\left(\frac{\pi }{6}\right)\end{array}$ .
     ::如果y=cscx,请找到y'(x)6。
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  10. $\begin{array}{r}y=x^5\cos (x)\end{array}$ ?
      ::y=x5cos(x)?
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  11. What is the derivative of $\begin{array}{r}{x}^2\csc (x)\end{array}$ ?
      ::x2csc(x) 的衍生物是什么?
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  12. What is the derivative of $\begin{array}{r}\csc (x)\tan (x)\end{array}$ ?
      ::Csc(x)tan(x)的衍生物是什么?
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  13. Use the quotient rule to verify that the derivative of  $\begin{array}{r}\sec (x)\end{array}$ is $\begin{array}{r}\sec (x)\tan (x)\end{array}$ .
      ::使用商数规则来核实 sec(x) 的衍生物是 sec(x)tan(x) 。
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  14. What is the derivative of $\begin{array}{r}\cot \left(\frac{\pi }{2}-x\right)\end{array}$ ?
      ::COTQQ( 2- x) 的衍生物是什么 ?
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  15. What is the derivative of $\begin{array}{r}{\csc }^2(x)-{\cot }^2(x)\end{array}$ ?
      ::Csc2(x)-cot2(x)的衍生物是什么?

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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 。