【高等数学学习记录】函数的求导法则

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一、知识点


(一)常数和基本初等函数的导数公式

  • ( C ) ′ = 0 (C)'=0 (C)=0
  • ( x n ) ′ = n x n − 1 (x^n)'=nx^{n-1} (xn)=nxn1
  • ( s i n x ) ′ = c o s x (sinx)'=cosx (sinx)=cosx
  • ( c o s x ) ′ = − s i n x (cosx)'=-sinx (cosx)=sinx
  • ( t a n x ) ′ = s e c 2 x (tanx)'=sec^2x (tanx)=sec2x
  • ( c o t x ) ′ = − c s c 2 x (cotx)'=-csc^2x (cotx)=csc2x
  • ( s e c x ) ′ = s e c x ⋅ t a n x (secx)'=secx\cdot tanx (secx)=secxtanx
  • ( c s c x ) ′ = − c s c x ⋅ c o t x (cscx)'=-cscx\cdot cotx (cscx)=cscxcotx
  • ( a x ) ′ = a x l n a (a^x)'=a^xlna (ax)=axlna
  • ( e x ) ′ = e x (e^x)'=e^x (ex)=ex
  • ( l o g a x ) ′ = 1 x l n a (log_ax)'=\frac{1}{xlna} (logax)=xlna1
  • ( l n x ) ′ = 1 x (lnx)'=\frac{1}{x} (lnx)=x1
  • ( a r c s i n x ) ′ = 1 1 − x 2 (arcsinx)'=\frac{1}{\sqrt{1-x^2}} (arcsinx)=1x2 1
  • ( a r c c o s x ) ′ = − 1 1 − x 2 (arccosx)'=-\frac{1}{\sqrt{1-x^2}} (arccosx)=1x2 1
  • ( a r c t a n x ) ′ = 1 1 + x 2 (arctanx)'=\frac{1}{1+x^2} (arctanx)=1+x21
  • ( a r c c o t x ) ′ = − 1 1 + x 2 (arccotx)'=-\frac{1}{1+x^2} (arccotx)=1+x21

(二)函数的和、差、积、商的求导法则

  • u = u ( x ) , v = v ( x ) u=u(x),v=v(x) u=u(x),v=v(x) 都可导,则
  • ( u ± v ) ′ = u ′ ± v ′ (u\pm v)'=u'\pm v' (u±v)=u±v
  • ( C u ) ′ = C u ′ (Cu)'=Cu' (Cu)=Cu C C C 是常数)
  • ( u v ) ′ = u ′ v + u v ′ (uv)'=u'v+uv' (uv)=uv+uv
  • ( u v ) ′ = u ′ v + u v ′ v 2 ( v ≠ 0 ) (\frac{u}{v})'=\frac{u'v+uv'}{v^2}(v\neq 0) (vu)=v2uv+uv(v=0)

(三)反函数的求导法则

  • x = f ( y ) x=f(y) x=f(y) 在区间 I y I_y Iy 内单调、可导且 f ′ ( y ) ≠ 0 f'(y)\neq 0 f(y)=0,则它的反函数 y = f − 1 ( x ) y=f^{-1}(x) y=f1(x) I x = f ( I y ) I_x=f(I_y) Ix=f(Iy) 内也可导,且 [ f − 1 ( x ) ] ′ = 1 f ′ ( y ) [f^{-1}(x)]'=\frac{1}{f'(y)} [f1(x)]=f(y)1 d y d x = 1 d x d y \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}} dxdy=dydx1.

(四)复合函数的求导法则

  • y = f ( u ) y=f(u) y=f(u),而 u = g ( x ) u=g(x) u=g(x) f ( u ) f(u) f(u) g ( x ) g(x) g(x) 都可导,则复合函数 y = f [ g ( x ) ] y=f[g(x)] y=f[g(x)] 的导数为 d y d x = d y d u ⋅ d u d x \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx} dxdy=dudydxdu y ′ ( x ) = f ′ ( u ) ⋅ g ′ ( x ) y'(x)=f'(u)\cdot g'(x) y(x)=f(u)g(x).

二、练习题


  1. 推导余切函数及余割函数的导数公式: ( c o t x ) ′ = − c s c 2 x (cotx)'=-csc^2x (cotx)=csc2x ( c s c x ) ′ = − c s c x ⋅ c o t x (cscx)'=-cscx\cdot cotx (cscx)=cscxcotx.
  • 推导:
  • (1) 推导余切函数的导数公式:
    ( c o t x ) ′ (cotx)' (cotx)
    = ( c o s x s i n x ) ′ =(\frac{cosx}{sinx})' =(sinxcosx)
    = ( c o s x ) ′ s i n x − c o s x ( s i n x ) ′ s i n 2 x =\frac{(cosx)'sinx-cosx(sinx)'}{sin^2x} =sin2x(cosx)sinxcosx(sinx)
    = − s i n 2 x − c o s 2 x s i n 2 x =\frac{-sin^2x-cos^2x}{sin^2x} =sin2xsin2xcos2x
    = − 1 s i n 2 x =-\frac{1}{sin^2x} =sin2x1
    = − c s c 2 x =-csc^2x =csc2x
  • (2) 推导余割函数的导数公式:
    ( c s c x ) ′ (cscx)' (cscx)
    = ( 1 s i n x ) ′ =(\frac{1}{sinx})' =(sinx1)
    = − c o s x s i n 2 x =\frac{-cosx}{sin^2x} =sin2xcosx
    = − c s c x ⋅ c o t x =-cscx\cdot cotx =cscxcotx.

  1. 求下列函数的导数:
  • (1) y = x 3 + 7 x 4 − 2 x + 12 y=x^3+\frac{7}{x^4}-\frac{2}{x}+12 y=x3+x47x2+12
    y ′ = 3 x 2 − 28 x 5 + 2 x 2 y'=3x^2-\frac{28}{x^5}+\frac{2}{x^2} y=3x2x528+x22
  • (2) y = 5 x 3 − 2 x + 3 e x y=5x^3-2^x+3e^x y=5x32x+3ex
    y ′ = 15 x 2 − 2 x l n 2 + 3 e x y'=15x^2-2^xln2+3e^x y=15x22xln2+3ex
  • (3) y = 2 t a n x + s e c x − 1 y=2tanx+secx-1 y=2tanx+secx1
    y ′ = 2 s e c 2 x + s e c x ⋅ t a n x y'=2sec^2x+secx\cdot tanx y=2sec2x+secxtanx
  • (4) y = s i n x ⋅ c o s x y=sinx\cdot cosx y=sinxcosx
    y ′ = c o s 2 x − s i n 2 x = c o s 2 x y'=cos^2x-sin^2x=cos2x y=cos2xsin2x=cos2x

  1. 求下列函数在给定点处的导数。
  • (1)
    y = s i n x − c o s x y=sinx - cosx y=sinxcosx,求 y ′ ∣ x = π 6 y'|_{x=\frac{\pi}{6}} yx=6π y ′ ∣ x = π 4 y'|_{x=\frac{\pi}{4}} yx=4π
    y ′ = c o s x + s i n x y'=cosx+sinx y=cosx+sinx
    y ′ ∣ x = π 6 = c o s π 6 + s i n π 6 = 3 2 + 1 2 = 3 + 1 2 y'|_{x=\frac{\pi}{6}}=cos\frac{\pi}{6}+sin\frac{\pi}{6}=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2} yx=6π=cos6π+sin6π=23 +21=23 +1
    y ′ ∣ x = π 4 = c o s π 4 + s i n π 4 = 2 2 + 2 2 = 2 y'|_{x=\frac{\pi}{4}}=cos{\frac{\pi}{4}}+sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2} yx=4π=cos4π+sin4π=22 +22 =2

  • (2)
    ρ = θ s i n θ + 1 2 c o s θ \rho = \theta sin\theta+\frac{1}{2}cos\theta ρ=θsinθ+21cosθ,求 d ρ d θ ∣ θ = π 4 \frac{d\rho}{d\theta}|_{\theta=\frac{\pi}{4}} dθdρθ=4π
    d ρ d θ = s i n θ + θ c o s θ − 1 2 s i n θ = θ c o s θ + 1 2 s i n θ \frac{d\rho}{d\theta}=sin\theta+\theta cos\theta -\frac{1}{2}sin\theta=\theta cos\theta +\frac{1}{2}sin\theta dθdρ=sinθ+θcosθ21sinθ=θcosθ+21sinθ
    d ρ d θ ∣ θ = π 4 = θ c o s θ + 1 2 s i n θ ∣ θ = π 4 = π 4 ⋅ 2 2 + 1 2 ⋅ 2 2 = 2 ( π + 2 ) 8 \frac{d\rho}{d\theta}|_{\theta = \frac{\pi}{4}}=\theta cos\theta + \frac{1}{2}sin\theta|_{\theta =\frac{\pi}{4}}=\frac{\pi}{4}\cdot \frac{\sqrt{2}}{2}+\frac{1}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}(\pi+2)}{8} dθdρθ=4π=θcosθ+21sinθθ=4π=4π22 +2122 =82 (π+2)

  • (3)
    f ( x ) = 3 5 − x + x 2 5 f(x)=\frac{3}{5-x}+\frac{x^2}{5} f(x)=5x3+5x2,求 f ′ ( 0 ) f'(0) f(0) f ′ ( 2 ) f'(2) f(2).
    f ′ ( x ) = 3 ⋅ ( − 1 ) ⋅ ( 5 − x ) − 2 ⋅ ( − 1 ) + 2 5 x = 3 ( 5 − x ) 2 + 2 5 x f'(x)=3\cdot (-1)\cdot (5-x)^{-2}\cdot (-1)+\frac{2}{5}x=\frac{3}{(5-x)^2}+\frac{2}{5}x f(x)=3(1)(5x)2(1)+52x=(5x)23+52x
    f ′ ( 0 ) = 3 25 f'(0)=\frac{3}{25} f(0)=253
    f ′ ( 2 ) = − 3 9 + 4 5 = 17 15 f'(2)=-\frac{3}{9}+\frac{4}{5}=\frac{17}{15} f(2)=93+54=1517


  1. 以初速度 v 0 v_0 v0 竖直上抛的物体,其上升高度 s s s 与世间 t t t 的关系是 s = v 0 t − 1 2 g t 2 s=v_0t-\frac{1}{2}gt^2 s=v0t21gt2. 求:
    (1) 该物体的速度 v ( t ) v(t) v(t)
    (2) 该物体达到最高点的时刻.
  • 解答:
  • (1)
    v ( t ) = s ′ ( t ) = v 0 − g t v(t)=s'(t)=v_0-gt v(t)=s(t)=v0gt
  • (2)
    该物体达到最高点时的速度为0, 即: v 0 − g t = 0 v_0-gt=0 v0gt=0
    t = v 0 g t=\frac{v_0}{g} t=gv0.

  1. 求曲线 y = 2 s i n x + x 2 y=2sinx+x^2 y=2sinx+x2 上横坐标为 x = 0 x=0 x=0 的点处的切线方程和法线方程.
  • 解答:
    根据曲线方程,当横坐标 x = 0 x=0 x=0 时,纵坐标 y = 0 y=0 y=0.
    x = 0 x=0 x=0 处的切线斜率为 y ′ ∣ x = 0 = ( 2 c o s x + 2 x ) ∣ x = 0 = 2 y'|_{x=0}=(2cosx+2x)|_{x=0}=2 yx=0=(2cosx+2x)x=0=2
    求得切线方程为 y − 0 = 2 ( x − 0 ) y-0=2(x-0) y0=2(x0),即 y = 2 x y=2x y=2x
    x = 0 x=0 x=0 处的法线斜率为 − 1 2 -\frac{1}{2} 21
    求得法线方程为 y − 0 = − 1 2 ( x − 0 ) y-0=-\frac{1}{2}(x-0) y0=21(x0),即 y = − 1 2 x y=-\frac{1}{2}x y=21x.

  1. 求下列函数的导数:
  • (1) y = ( 2 x + 5 ) 4 y=(2x+5)^4 y=(2x+5)4
    y ′ = 4 ⋅ ( 2 x + 5 ) 3 ⋅ 2 = 8 ⋅ ( 2 x + 5 ) 3 y'=4\cdot (2x+5)^3\cdot 2=8\cdot (2x+5)^3 y=4(2x+5)32=8(2x+5)3
  • (2) y = c o s ( 4 − 3 x ) y=cos(4-3x) y=cos(43x)
    y ′ = − s i n ( 4 − 3 x ) ⋅ ( − 3 ) = 3 ⋅ s i n ( 4 − 3 x ) y'=-sin(4-3x)\cdot (-3)=3\cdot sin(4-3x) y=sin(43x)(3)=3sin(43x)
  • (3) y = e − 3 x 2 y=e^{-3x^2} y=e3x2
    y ′ = e − 3 x 2 ⋅ ( − 3 ⋅ 2 ) ⋅ x = − 6 ⋅ x ⋅ e − 3 x 2 y'=e^{-3x^2}\cdot (-3\cdot 2)\cdot x=-6\cdot x\cdot e^{-3x^2} y=e3x2(32)x=6xe3x2
  • (4) y = l n ( 1 + x 2 ) y=ln(1+x^2) y=ln(1+x2)
    y ′ = 1 1 + x 2 ⋅ ( 0 + 2 x ) = 2 x 1 + x 2 y'=\frac{1}{1+x^2}\cdot (0+2x)=\frac{2x}{1+x^2} y=1+x21(0+2x)=1+x22x

  1. 求下列函数的导数:
  • (1) y = a r c s i n ( 1 − 2 x ) y=arcsin(1-2x) y=arcsin(12x)
    y ′ = 1 1 − ( 1 − 2 x ) 2 ( 1 − 2 x ) ′ = − 2 4 x − 4 x 2 = − 1 x − x 2 y'=\frac{1}{\sqrt{1-(1-2x)^2}}(1-2x)'=\frac{-2}{\sqrt{4x-4x^2}}=-\frac{1}{\sqrt{x-x^2}} y=1(12x)2 1(12x)=4x4x2 2=xx2 1
  • (2) y = 1 1 − x 2 y=\frac{1}{\sqrt{1-x^2}} y=1x2 1
    y ′ = − 1 2 ( 1 − x 2 ) − 3 2 ( 1 − x 2 ) ′ = − 1 2 ( 1 − x 2 ) − 3 2 ( − 2 x ) = x ( 1 − x 2 ) − 3 2 y'=-\frac{1}{2}(1-x^2)^{-\frac{3}{2}}(1-x^2)'=-\frac{1}{2}(1-x^2)^{-\frac{3}{2}}(-2x)=x(1-x^2)^{-\frac{3}{2}} y=21(1x2)23(1x2)=21(1x2)23(2x)=x(1x2)23
  • (3) y = e − x 2 c o s 3 x y=e^{-\frac{x}{2}}cos3x y=e2xcos3x
    y ′ = ( e − x 2 ) ′ c o s 3 x + e − 3 2 ( c o s 3 x ) ′ y'=(e^{-\frac{x}{2}})'cos3x+e^{-\frac{3}{2}}(cos3x)' y=(e2x)cos3x+e23(cos3x)
    = − 1 2 e − x 2 c o s 3 x − 3 e − 3 2 s i n 3 x =-\frac{1}{2}e^{-\frac{x}{2}}cos3x-3e^{-\frac{3}{2}}sin3x =21e2xcos3x3e23sin3x
  • (4) y = a r c c o s 1 x y=arccos\frac{1}{x} y=arccosx1
    y ′ = − 1 1 − 1 x 2 ⋅ ( − 1 x 2 ) y'=-\frac{1}{\sqrt{1-\frac{1}{x^2}}}\cdot (-\frac{1}{x^2}) y=1x21 1(x21)
    = 1 ∣ x ∣ x 2 − 1 =\frac{1}{|x|\sqrt{x^2-1}} =xx21 1

  1. 求下列函数的导数:
  • (1) y = ( a r c s i n x 2 ) 2 y=(arcsin\frac{x}{2})^2 y=(arcsin2x)2
    y ′ = 2 ⋅ a r c s i n x 2 ⋅ ( a r c s i n x 2 ) ′ y'=2\cdot arcsin\frac{x}{2}\cdot (arcsin\frac{x}{2})' y=2arcsin2x(arcsin2x)
    = 2 ⋅ a r c s i n x 2 ⋅ 1 1 − ( x 2 ) 2 ⋅ ( x 2 ) ′ =2\cdot arcsin\frac{x}{2}\cdot \frac{1}{\sqrt{1-(\frac{x}{2})^2}}\cdot (\frac{x}{2})' =2arcsin2x1(2x)2 1(2x)
    = 2 ⋅ a r c s i n x 2 4 − x 2 =\frac{2\cdot arcsin\frac{x}{2}}{\sqrt{4-x^2}} =4x2 2arcsin2x
  • (2) y = l n t a n x 2 y=ln tan\frac{x}{2} y=lntan2x
    y ′ = 1 t a n x 2 ( t a n x 2 ) ′ y'=\frac{1}{tan\frac{x}{2}}(tan\frac{x}{2})' y=tan2x1(tan2x)
    = 1 t a n x 2 ⋅ s e c 2 x 2 ⋅ ( x 2 ) ′ =\frac{1}{tan\frac{x}{2}}\cdot sec^2\frac{x}{2}\cdot (\frac{x}{2})' =tan2x1sec22x(2x)
    = 1 2 ⋅ s i n x 2 ⋅ c o s x 2 =\frac{1}{2\cdot sin\frac{x}{2}\cdot cos\frac{x}{2}} =2sin2xcos2x1
    = 1 s i n x = c s c x =\frac{1}{sinx}=cscx =sinx1=cscx
  • (3) y = 1 + l n 2 x y=\sqrt{1+ln^2x} y=1+ln2x
    y ′ = 1 2 1 + l n 2 x ⋅ ( 1 + l n 2 x ) ′ y'=\frac{1}{2\sqrt{1+ln^2x}}\cdot (1+ln^2x)' y=21+ln2x 1(1+ln2x)
    = 1 2 ⋅ 1 + l n 2 x ⋅ 2 ⋅ l n x ⋅ 1 x =\frac{1}{2\cdot \sqrt{1+ln^2x}}\cdot 2\cdot lnx\cdot \frac{1}{x} =21+ln2x 12lnxx1
    = l n x x 1 + l n 2 x =\frac{lnx}{x\sqrt{1+ln^2x}} =x1+ln2x lnx
  • (4) y = e a r c t a n x y=e^{arctan\sqrt{x}} y=earctanx
    y ′ = e a r c t a n x ⋅ ( a r c t a n x y'=e^{arctan\sqrt{x}}\cdot (arctan\sqrt{x} y=earctanx (arctanx )’
    = e a r c t a n x ⋅ 1 1 + x ⋅ ( x ) ′ =e^{arctan\sqrt{x}}\cdot \frac{1}{1+x}\cdot (\sqrt{x})' =earctanx 1+x1(x )
    = e a r c t a n x 2 ⋅ ( 1 + x ) ⋅ x =\frac{e^{arctan\sqrt{x}}}{2\cdot (1+x)\cdot \sqrt{x}} =2(1+x)x earctanx

  1. 设函数 f ( x ) f(x) f(x) g ( x ) g(x) g(x) 可导,且 f 2 ( x ) + g 2 ( x ) ≠ 0 f^2(x)+g^2(x)\neq 0 f2(x)+g2(x)=0,试求函数 y = f 2 ( x ) + g 2 ( x ) y=\sqrt{f^2(x)+g^2(x)} y=f2(x)+g2(x) 的导数.
  • 解:
    y ′ = ( f 2 ( x ) + g 2 ( x ) ) ′ y'=(\sqrt{f^2(x)+g^2(x)})' y=(f2(x)+g2(x) )
    = 1 2 f 2 ( x ) + g 2 ( x ) ⋅ [ f 2 ( x ) + g 2 ( x ) ] ′ =\frac{1}{2\sqrt{f^2(x)+g^2(x)}}\cdot [f^2(x)+g^2(x)]' =2f2(x)+g2(x) 1[f2(x)+g2(x)]
    = 2 f ( x ) f ′ ( x ) + 2 g ( x ) g ′ ( x ) 2 f 2 ( x ) + g 2 ( x ) =\frac{2f(x)f'(x)+2g(x)g'(x)}{2\sqrt{f^2(x)+g^2(x)}} =2f2(x)+g2(x) 2f(x)f(x)+2g(x)g(x)
    = f ( x ) f ′ ( x ) + g ( x ) g ′ ( x ) f 2 ( x ) + g 2 ( x ) =\frac{f(x)f'(x)+g(x)g'(x)}{\sqrt{f^2(x)+g^2(x)}} =f2(x)+g2(x) f(x)f(x)+g(x)g(x)

  1. 设函数 f ( x ) f(x) f(x) 可导,求下列函数的导数 d y d x \frac{dy}{dx} dxdy

    (1) y = f ( x 2 ) y=f(x^2) y=f(x2)

    (2) y = f ( s i n 2 x ) + f ( c o s 2 x ) y=f(sin^2x)+f(cos^2x) y=f(sin2x)+f(cos2x)

  • 解:

  • (1)

    d y d x = d ( f ( x 2 ) ) d x = f ′ ( x 2 ) ⋅ ( x 2 ) ′ = 2 x ⋅ f ′ ( x 2 ) \frac{dy}{dx}=\frac{d(f(x^2))}{dx}=f'(x^2)\cdot (x^2)'=2x\cdot f'(x^2) dxdy=dxd(f(x2))=f(x2)(x2)=2xf(x2)

  • (2)

    d y d x = f ′ ( s i n 2 x ) ⋅ ( s i n 2 x ) ′ + f ′ ( c o s 2 x ) ⋅ ( c o s 2 x ) ′ \frac{dy}{dx}=f'(sin^2x)\cdot (sin^2x)'+f'(cos^2x)\cdot (cos^2x)' dxdy=f(sin2x)(sin2x)+f(cos2x)(cos2x)

    = 2 s i n x c o s x f ′ ( s i n 2 x ) + 2 c o s x ( − s i n x ) f ′ ( c o s 2 x ) =2sinxcosxf'(sin^2x)+2cosx(-sinx)f'(cos^2x) =2sinxcosxf(sin2x)+2cosx(sinx)f(cos2x)

    = 2 s i n x c o s x [ f ′ ( s i n 2 x ) − f ′ ( c o s 2 x ) ] =2sinxcosx[f'(sin^2x)-f'(cos^2x)] =2sinxcosx[f(sin2x)f(cos2x)]

    = s i n 2 x [ f ′ ( s i n 2 x ) − f ′ ( c o s 2 x ) ] =sin2x[f'(sin^2x)-f'(cos^2x)] =sin2x[f(sin2x)f(cos2x)]


  1. 求下列函数的导数:
  • (1) y = e − x ( x 2 − 2 x + 3 ) y=e^{-x}(x^2-2x+3) y=ex(x22x+3)

    y ′ = ( e − x ) ′ ⋅ ( x 2 − 2 x + 3 ) + e − x ⋅ ( x 2 − 2 x + 3 ) ′ y'=(e^{-x})'\cdot (x^2-2x+3)+e^{-x}\cdot (x^2-2x+3)' y=(ex)(x22x+3)+ex(x22x+3)

    = e − x ⋅ ( − 1 ) ⋅ ( x 2 − 2 x + 3 ) + e − x ⋅ ( 2 x − 2 ) =e^{-x}\cdot (-1)\cdot (x^2-2x+3)+e^{-x}\cdot (2x-2) =ex(1)(x22x+3)+ex(2x2)

    = e − x ( − x 2 + 2 x − 3 + 2 x − 2 ) =e^{-x}(-x^2+2x-3+2x-2) =ex(x2+2x3+2x2)

    = e − x ( 4 x − x 2 − 5 ) =e^{-x}(4x-x^2-5) =ex(4xx25)

  • (2) y = s i n 2 x ⋅ s i n ( x 2 ) y=sin^2x\cdot sin(x^2) y=sin2xsin(x2)

    y ′ = ( s i n 2 x ) ′ ⋅ s i n ( x 2 ) + s i n 2 ⋅ [ s i n ( x 2 ) ] ′ y'=(sin^2x)'\cdot sin(x^2)+sin^2\cdot [sin(x^2)]' y=(sin2x)sin(x2)+sin2[sin(x2)]

    = 2 ⋅ s i n x ⋅ c o s x ⋅ s i n ( x 2 ) + s i n 2 ⋅ c o s ( x 2 ) ⋅ 2 ⋅ x =2\cdot sinx\cdot cosx\cdot sin(x^2)+sin^2\cdot cos(x^2)\cdot 2\cdot x =2sinxcosxsin(x2)+sin2cos(x2)2x

    = s i n 2 x ⋅ s i n ( x 2 ) + 2 ⋅ x ⋅ s i n 2 x ⋅ c o s ( x 2 ) =sin2x\cdot sin(x^2)+2\cdot x\cdot sin^2x\cdot cos(x^2) =sin2xsin(x2)+2xsin2xcos(x2)

  • (3) y = ( a r c t a n x 2 ) 2 y=(arctan\frac{x}{2})^2 y=(arctan2x)2

    y ′ = 2 ⋅ a r c t a n x 2 ⋅ ( a r c t a n x 2 ) ′ y'=2\cdot arctan\frac{x}{2}\cdot (arctan\frac{x}{2})' y=2arctan2x(arctan2x)

    = 2 ⋅ a r c t a n x 2 ⋅ 1 1 + ( x 2 ) 2 ⋅ ( x 2 ) ′ =2\cdot arctan\frac{x}{2}\cdot \frac{1}{1+(\frac{x}{2})^2}\cdot (\frac{x}{2})' =2arctan2x1+(2x)21(2x)

    = 2 ⋅ a r c t a n x 2 ⋅ 1 1 + x 2 4 ⋅ 1 2 =2\cdot arctan\frac{x}{2}\cdot \frac{1}{1+\frac{x^2}{4}}\cdot \frac{1}{2} =2arctan2x1+4x2121

    = 4 a r c t a n x 2 4 + x 2 =\frac{4arctan\frac{x}{2}}{4+x^2} =4+x24arctan2x

  • (4) y = l n x x n y=\frac{lnx}{x^n} y=xnlnx

    y ′ = ( l n x ) ′ x n − l n x ( x n ) ′ x 2 n y'=\frac{(lnx)'x^n-lnx(x^n)'}{x^{2n}} y=x2n(lnx)xnlnx(xn)

    = 1 x ⋅ x n − l n x ⋅ n x n − 1 x 2 n =\frac{\frac{1}{x}\cdot x^n-lnx\cdot n x^{n-1}}{x^{2n}} =x2nx1xnlnxnxn1

    = 1 − n l n x x n + 1 =\frac{1-nlnx}{x^{n+1}} =xn+11nlnx


  1. 设函数 f ( x ) f(x) f(x) g ( x ) g(x) g(x) 均在点 x 0 x_0 x0 的某一邻域内有定义, f ( x ) f(x) f(x) x 0 x_0 x0 处可导, f ( x 0 ) = 0 f(x_0)=0 f(x0)=0 g ( x ) g(x) g(x) x 0 x_0 x0 处连续,试讨论 f ( x ) g ( x ) f(x)g(x) f(x)g(x) x 0 x_0 x0 处的可导性.
  • 解:

    ∵ f ( x ) \because f(x) f(x) x 0 x_0 x0 处可导,且 f ( x 0 ) = 0 f(x_0)=0 f(x0)=0

    ∴ f ′ ( x 0 ) = lim ⁡ x → x 0 f ( x ) − f ( x 0 ) x − x 0 = lim ⁡ x → x 0 f ( x ) x − x 0 \therefore f'(x_0)=\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\rightarrow x_0}\frac{f(x)}{x-x_0} f(x0)=limxx0xx0f(x)f(x0)=limxx0xx0f(x)

    ∵ g ( x ) \because g(x) g(x) x 0 x_0 x0 处连续

    ∴ lim ⁡ x → x 0 g ( x ) = g ( x 0 ) \therefore \lim_{x\rightarrow x_0}g(x)=g(x_0) limxx0g(x)=g(x0)

    ∴ lim ⁡ x → x 0 f ( x ) g ( x ) − f ( x 0 ) g ( x 0 ) x − x 0 \therefore \lim_{x\rightarrow x_0}\frac{f(x)g(x)-f(x_0)g(x_0)}{x-x_0} limxx0xx0f(x)g(x)f(x0)g(x0)

    = lim ⁡ x → x 0 f ( x ) x − x 0 g ( x ) = f ′ ( x 0 ) g ( x 0 ) =\lim_{x\rightarrow x_0}\frac{f(x)}{x-x_0}g(x)=f'(x_0)g(x_0) =limxx0xx0f(x)g(x)=f(x0)g(x0)

    ∴ f ( x ) g ( x ) \therefore f(x)g(x) f(x)g(x) x 0 x_0 x0 处可导.


  1. 设函数 f ( x ) f(x) f(x) 满足下列条件:

    (1) f ( x + y ) = f ( x ) ⋅ f ( y ) f(x+y)=f(x)\cdot f(y) f(x+y)=f(x)f(y),对一切 x , y ∈ R x,y\in R x,yR

    (2) f ( x ) = 1 + x g ( x ) f(x)=1+xg(x) f(x)=1+xg(x),而 lim ⁡ x → 0 g ( x ) = 1 \lim_{x\rightarrow 0}g(x)=1 limx0g(x)=1

    试证明 f ( x ) f(x) f(x) R R R 上处处可导,且 f ′ ( x ) = f ( x ) f'(x)=f(x) f(x)=f(x).

  • 证明:

    f ′ ( x ) = lim ⁡ Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x f'(x)=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} f(x)=limΔx0Δxf(x+Δx)f(x)

    = lim ⁡ Δ x → 0 f ( x ) f ( Δ x ) − f ( x ) Δ x =\lim_{\Delta x\rightarrow 0}\frac{f(x)f(\Delta x)-f(x)}{\Delta x} =limΔx0Δxf(x)f(Δx)f(x)

    = lim ⁡ Δ x → 0 [ f ( x ) ⋅ f ( Δ x ) − 1 Δ x ] =\lim_{\Delta x\rightarrow 0}[f(x)\cdot \frac{f(\Delta x)-1}{\Delta x}] =limΔx0[f(x)Δxf(Δx)1]

    = lim ⁡ Δ x → 0 [ f ( x ) ⋅ Δ x g ( Δ x ) Δ x ] =\lim_{\Delta x\rightarrow 0}[f(x)\cdot \frac{\Delta xg(\Delta x)}{\Delta x}] =limΔx0[f(x)ΔxΔxg(Δx)]

    = lim ⁡ Δ x → 0 [ f ( x ) g ( Δ x ) ] =\lim_{\Delta x\rightarrow 0}[f(x)g(\Delta x)] =limΔx0[f(x)g(Δx)]

    = f ( x ) =f(x) =f(x)



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