3.4 Derivatives of Trigonometric Functions

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三角函数的导数可以从sin(x)和cos(x)两个函数出发，推导出其它的三角函数的导数。因此这两个函数的导数求法很重要。


ddxsin(x)=cos(x)\frac{d}{dx}sin(x) = cos(x)

证明：

f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0sin⁡(x+h)−sin⁡xh=lim⁡h→0(sin⁡xcos⁡h+cos⁡xsin⁡h)−sin⁡xh=lim⁡h→0sin⁡x(cos⁡h−1)+cos⁡xsin⁡hh=lim⁡h→0(sin⁡x⋅cos⁡h−1h)+lim⁡h→0(cos⁡x⋅sin⁡hh)=sin⁡x⋅lim⁡h→0cos⁡h−1h+cos⁡x⋅lim⁡h→0sin⁡hh=sin⁡x⋅0+cos⁡x⋅1=cos⁡x\begin{aligned} f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \to 0}\frac{\sin(x + h) - \sin x}{h} \\ &= \lim_{h \to 0}\frac{(\sin x \cos h + \cos x \sin h) - \sin x}{h} \\ &= \lim_{h \to 0}\frac{\sin x(\cos h - 1) + \cos x \sin h}{h} \\ &= \lim_{h \to 0}(\sin x \cdot \frac{\cos h - 1}{h}) + \lim_{h \to 0}(\cos x \cdot \frac{\sin h}{h}) \\ &= \sin x \cdot \lim_{h \to 0}\frac{\cos h - 1}{h} + \cos x \cdot \lim_{h \to 0}\frac{\sin h}{h} \\ &= \sin x \cdot 0 + \cos x \cdot 1 \\ &= \cos x \end{aligned}

ddxcos(x)=−sin(x)\frac{d}{dx}cos(x) = -sin(x)

证明：

ddx(cos⁡x)=lim⁡h→0cos⁡(x+h)−cos⁡xh=lim⁡h→0(cos⁡xcos⁡h−sin⁡xsin⁡h)−cos⁡xh=lim⁡h→0cos⁡x(cos⁡h−1)−sin⁡xsin⁡hh=cos⁡x⋅lim⁡h→0(cos⁡h−1h)−sin⁡x⋅lim⁡h→0sin⁡hh=cos⁡x⋅0−sin⁡x⋅1=−sin⁡x\begin{aligned} \frac{d}{dx}(\cos x) &= \lim_{h \to 0}\frac{\cos(x + h) - \cos x}{h} \\ &= \lim_{h \to 0}\frac{(\cos x \cos h - \sin x \sin h) - \cos x}{h} \\ &= \lim_{h \to 0}\frac{\cos x (\cos h - 1) - \sin x \sin h}{h} \\ &= \cos x \cdot \lim_{h \to 0}(\frac{\cos h - 1}{h}) - \sin x \cdot \lim_{h \to 0}\frac{\sin h}{h} \\ &= \cos x \cdot 0 - \sin x \cdot 1 \\ &= -\sin x \end{aligned}

由此，根据tan(x)，sec(x)，cot(x)和csc(x)的定义，可以推导出他们也是可导的。

ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x

ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x) = \sec x \tan x

ddx(cot⁡x)=−csc⁡2x\frac{d}{dx}(\cot x) = -\csc^2 x

ddx(csc⁡x)=−csc⁡xcot⁡x\frac{d}{dx}(\csc x) = -\csc x \cot x


这些三角函数可导，引出一个推论就是，它们都是连续的。



练习

  1. y=−10x+3cos⁡xy = -10x + 3\cos x

dydx=−10−3sin⁡x\frac{dy}{dx} = -10 - 3\sin x


  1. y=3x+5sin⁡xy = \frac{3}{x} + 5\sin x

dydx=−32x2+5cos⁡x\frac{dy}{dx} = -\frac{3}{2x^2} + 5\cos x


  1. y=csc⁡x−4x+7y = \csc x - 4\sqrt{x} + 7

y′=−csc⁡xcot⁡x−2xy' = -\csc x \cot x -\frac{2}{\sqrt{x}}


  1. y=x2cot⁡x−1x2y = x^2\cot x - \frac{1}{x^2}

y′=2xcot⁡x−x2csc⁡2x+2x3y' = 2x\cot x - x^2\csc^2 x + \frac{2}{x^3}


  1. y=(sec⁡x+tan⁡x)(sec⁡x−tan⁡x)y = (\sec x + \tan x)(\sec x - \tan x)

y′=ddx(sec⁡2x−tan⁡2x)=2sec⁡2xtan⁡x−2tan⁡xsec⁡2x=0y' = \frac{d}{dx}(\sec^2 x - \tan^2 x) = 2\sec^2 x \tan x - 2\tan x \sec^2 x = 0


  1. y=(sin⁡x+cos⁡x)sec⁡xy = (\sin x + \cos x)\sec x

y′=(cos⁡x−sin⁡x)sec⁡x+(sin⁡x+cos⁡x)sec⁡xtan⁡x=1+tan⁡2xy' = (\cos x - \sin x)\sec x + (\sin x + \cos x)\sec x \tan x = 1 + \tan^2 x


  1. y=cot⁡x1+cot⁡xy = \frac{\cot x}{1 + \cot x}

y′=−csc⁡2x(1+cot⁡x)+cot⁡xcsc⁡2x(1+cot⁡x)2y' = \frac{-\csc^2 x(1 + \cot x) + \cot x \csc^2 x}{(1 + \cot x)^2}


  1. y=cos⁡x1+sin⁡xy = \frac{\cos x}{1 + \sin x}

y′=−sin⁡x(1+sin⁡x)−cos⁡2x(1+sin⁡x)2=−1sin⁡xy' = \frac{-\sin x(1 + \sin x) - \cos^2 x}{(1 + \sin x)^2} = -\frac{1}{\sin x}


  1. y=4cos⁡x+1tan⁡xy = \frac{4}{\cos x} + \frac{1}{\tan x}

y′=ddx(4sec⁡x+cot⁡x)=4sec⁡xtan⁡x−csc⁡2xy' = \frac{d}{dx}(4\sec x + \cot x) = 4\sec x \tan x - \csc^2 x


  1. y=cos⁡xx+xcos⁡xy = \frac{\cos x}{x} + \frac{x}{\cos x}

y′=−sin⁡x⋅x−cos⁡xx2+cos⁡x−sin⁡x⋅xcos⁡2x=−sin⁡xx−cos⁡xx2+1cos⁡x−xsin⁡xcos⁡2xy' = \frac{-\sin x \cdot x - \cos x}{x^2} + \frac{\cos x - \sin x \cdot x}{\cos^2 x} = -\frac{\sin x}{x} - \frac{\cos x}{x^2} + \frac{1}{\cos x} - \frac{x\sin x}{\cos^2 x}


  1. y=x2sin⁡x+2xcos⁡x−2sin⁡xy = x^2\sin x + 2x\cos x - 2\sin x

y′=2xsin⁡x+x2cos⁡x+2cos⁡x−2xsin⁡x−2cos⁡x=x2cos⁡xy' = 2x\sin x + x^2\cos x + 2\cos x - 2x\sin x - 2\cos x = x^2\cos x


  1. y=x2cos⁡x−2xsin⁡x−2cos⁡xy = x^2\cos x - 2x\sin x - 2\cos x

y′=2xcos⁡x−x2sin⁡x−2sin⁡x−2xcos⁡x+2sin⁡x=−x2sin⁡xy' = 2x\cos x - x^2\sin x - 2\sin x - 2x\cos x + 2\sin x = -x^2\sin x


  1. s=tan⁡t−ts = \tan t - t

s′=sec⁡2t−1s' = \sec^2 t - 1


  1. s=t2−sec⁡t+1s = t^2 - \sec t + 1

s′=2t−sec⁡ttan⁡ts' = 2t - \sec t \tan t


  1. s=1+csc⁡t1−csc⁡ts = \frac{1 + \csc t}{ 1 - \csc t}

s′=−csc⁡tcot⁡t+csc⁡tcot⁡t(1+csc⁡t)2=0s' = \frac{-\csc t \cot t + \csc t \cot t}{(1 + \csc t)^2} = 0


  1. s=sin⁡t1−cos⁡ts = \frac{\sin t}{1 - \cos t}

s′=cos⁡t(1−cos⁡t)−sin⁡2t(1−cos⁡t)2=−11−cos⁡ts' = \frac{\cos t(1 - \cos t) - \sin^2 t}{(1 - \cos t)^2} = -\frac{1}{1 - \cos t}

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