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)′=nxn−1
- ( 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)′=secx⋅tanx
- ( c s c x ) ′ = − c s c x ⋅ c o t x (cscx)'=-cscx\cdot cotx (cscx)′=−cscx⋅cotx
- ( 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)′=1−x21
- ( a r c c o s x ) ′ = − 1 1 − x 2 (arccosx)'=-\frac{1}{\sqrt{1-x^2}} (arccosx)′=−1−x21
- ( 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
- 设 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)′=u′v+uv′
- ( u v ) ′ = u ′ v + u v ′ v 2 ( v ≠ 0 ) (\frac{u}{v})'=\frac{u'v+uv'}{v^2}(v\neq 0) (vu)′=v2u′v+uv′(v=0)
(三)反函数的求导法则
- 设 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=f−1(x) 在 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)} [f−1(x)]′=f′(y)1 或 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
- 设 y = f ( u ) y=f(u) y=f(u),而 u = g ( x ) u=g(x) u=g(x) 且 f ( u ) f(u) f(u) 及 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=dudy⋅dxdu 或 y ′ ( x ) = f ′ ( u ) ⋅ g ′ ( x ) y'(x)=f'(u)\cdot g'(x) y′(x)=f′(u)⋅g′(x).
二、练习题
- 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)′=−cscx⋅cotx.
- 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)′sinx−cosx(sinx)′
= − s i n 2 x − c o s 2 x s i n 2 x =\frac{-sin^2x-cos^2x}{sin^2x} =sin2x−sin2x−cos2x
= − 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} =sin2x−cosx
= − c s c x ⋅ c o t x =-cscx\cdot cotx =−cscx⋅cotx.
- 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+x47−x2+12
y ′ = 3 x 2 − 28 x 5 + 2 x 2 y'=3x^2-\frac{28}{x^5}+\frac{2}{x^2} y′=3x2−x528+x22 - (2) y = 5 x 3 − 2 x + 3 e x y=5x^3-2^x+3e^x y=5x3−2x+3ex
y ′ = 15 x 2 − 2 x l n 2 + 3 e x y'=15x^2-2^xln2+3e^x y′=15x2−2xln2+3ex - (3) y = 2 t a n x + s e c x − 1 y=2tanx+secx-1 y=2tanx+secx−1
y ′ = 2 s e c 2 x + s e c x ⋅ t a n x y'=2sec^2x+secx\cdot tanx y′=2sec2x+secx⋅tanx - (4) y = s i n x ⋅ c o s x y=sinx\cdot cosx y=sinx⋅cosx
y ′ = c o s 2 x − s i n 2 x = c o s 2 x y'=cos^2x-sin^2x=cos2x y′=cos2x−sin2x=cos2x
- 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=sinx−cosx,求 y ′ ∣ x = π 6 y'|_{x=\frac{\pi}{6}} y′∣x=6π 和 y ′ ∣ x = π 4 y'|_{x=\frac{\pi}{4}} y′∣x=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} y′∣x=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} y′∣x=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}} dθdρ∣θ=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+θcosθ−21sinθ=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} dθdρ∣θ=4π=θcosθ+21sinθ∣θ=4π=4π⋅22+21⋅22=82(π+2) -
(3)
f ( x ) = 3 5 − x + x 2 5 f(x)=\frac{3}{5-x}+\frac{x^2}{5} f(x)=5−x3+5x2,求 f ′ ( 0 ) f'(0) f′(0) 和 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)⋅(5−x)−2⋅(−1)+52x=(5−x)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
- 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=v0t−21gt2. 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)=v0−gt - (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 v0−gt=0
得 t = v 0 g t=\frac{v_0}{g} t=gv0.
- 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 y′∣x=0=(2cosx+2x)∣x=0=2
求得切线方程为 y − 0 = 2 ( x − 0 ) y-0=2(x-0) y−0=2(x−0),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) y−0=−21(x−0),即 y = − 1 2 x y=-\frac{1}{2}x y=−21x.
- 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)3⋅2=8⋅(2x+5)3 - (2) y = c o s ( 4 − 3 x ) y=cos(4-3x) y=cos(4−3x)
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(4−3x)⋅(−3)=3⋅sin(4−3x) - (3) y = e − 3 x 2 y=e^{-3x^2} y=e−3x2
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′=e−3x2⋅(−3⋅2)⋅x=−6⋅x⋅e−3x2 - (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
- 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(1−2x)
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−(1−2x)21(1−2x)′=4x−4x2−2=−x−x21 - (2) y = 1 1 − x 2 y=\frac{1}{\sqrt{1-x^2}} y=1−x21
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(1−x2)−23(1−x2)′=−21(1−x2)−23(−2x)=x(1−x2)−23 - (3) y = e − x 2 c o s 3 x y=e^{-\frac{x}{2}}cos3x y=e−2xcos3x
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′=(e−2x)′cos3x+e−23(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 =−21e−2xcos3x−3e−23sin3x - (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′=−1−x211⋅(−x21)
= 1 ∣ x ∣ x 2 − 1 =\frac{1}{|x|\sqrt{x^2-1}} =∣x∣x2−11
- 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′=2⋅arcsin2x⋅(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})' =2⋅arcsin2x⋅1−(2x)21⋅(2x)′
= 2 ⋅ a r c s i n x 2 4 − x 2 =\frac{2\cdot arcsin\frac{x}{2}}{\sqrt{4-x^2}} =4−x22⋅arcsin2x - (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})' =tan2x1⋅sec22x⋅(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}} =2⋅sin2x⋅cos2x1
= 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+ln2x1⋅(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} =2⋅1+ln2x1⋅2⋅lnx⋅x1
= l n x x 1 + l n 2 x =\frac{lnx}{x\sqrt{1+ln^2x}} =x1+ln2xlnx - (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)⋅xearctanx
- 设函数 f ( x ) f(x) f(x) 和 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)
-
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)′=2x⋅f′(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)]
- 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=e−x(x2−2x+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′=(e−x)′⋅(x2−2x+3)+e−x⋅(x2−2x+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) =e−x⋅(−1)⋅(x2−2x+3)+e−x⋅(2x−2)
= e − x ( − x 2 + 2 x − 3 + 2 x − 2 ) =e^{-x}(-x^2+2x-3+2x-2) =e−x(−x2+2x−3+2x−2)
= e − x ( 4 x − x 2 − 5 ) =e^{-x}(4x-x^2-5) =e−x(4x−x2−5)
-
(2) y = s i n 2 x ⋅ s i n ( x 2 ) y=sin^2x\cdot sin(x^2) y=sin2x⋅sin(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 =2⋅sinx⋅cosx⋅sin(x2)+sin2⋅cos(x2)⋅2⋅x
= 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) =sin2x⋅sin(x2)+2⋅x⋅sin2x⋅cos(x2)
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(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′=2⋅arctan2x⋅(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})' =2⋅arctan2x⋅1+(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} =2⋅arctan2x⋅1+4x21⋅21
= 4 a r c t a n x 2 4 + x 2 =\frac{4arctan\frac{x}{2}}{4+x^2} =4+x24arctan2x
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(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)′xn−lnx(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}} =x2nx1⋅xn−lnx⋅nxn−1
= 1 − n l n x x n + 1 =\frac{1-nlnx}{x^{n+1}} =xn+11−nlnx
- 设函数 f ( x ) f(x) f(x) 和 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) 在 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) 在 x 0 x_0 x0 处连续,试讨论 f ( x ) g ( x ) f(x)g(x) f(x)g(x) 在 x 0 x_0 x0 处的可导性.
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解:
∵ f ( x ) \because f(x) ∵f(x) 在 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)=limx→x0x−x0f(x)−f(x0)=limx→x0x−x0f(x)
∵ g ( x ) \because g(x) ∵g(x) 在 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) ∴limx→x0g(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} ∴limx→x0x−x0f(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) =limx→x0x−x0f(x)g(x)=f′(x0)g(x0)
∴ f ( x ) g ( x ) \therefore f(x)g(x) ∴f(x)g(x) 在 x 0 x_0 x0 处可导.
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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,y∈R
(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 limx→0g(x)=1
试证明 f ( x ) f(x) f(x) 在 R R R 上处处可导,且 f ′ ( x ) = f ( x ) f'(x)=f(x) f′(x)=f(x).
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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)=limΔx→0Δxf(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} =limΔx→0Δxf(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}] =limΔx→0[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}] =limΔx→0[f(x)⋅ΔxΔxg(Δx)]
= lim Δ x → 0 [ f ( x ) g ( Δ x ) ] =\lim_{\Delta x\rightarrow 0}[f(x)g(\Delta x)] =limΔx→0[f(x)g(Δx)]
= f ( x ) =f(x) =f(x)
- 学习资料
1.《高等数学(第六版)》 上册,同济大学数学系 编
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