可推广到任意有限项的情形.
(2) (uv)′ = u′v + u v′
证: 设 f (x) = u(x)v(x) , 则有
f (x + h) − f (x) u(x + h)v(x + h) − u(x)v(x) f ′(x) = lim = lim h→0 h→0 h h
u(x + h) −u(x) v(x + h) + u(x) v(x + h) − v(x) = lim h→0 h h

例2. 求证
′ sin x (sin x)′ cos x − sin x (cos x)′ 证: (tan x)′ = = cos 2 x cos x
cos x + sin2 x = sec2 x = 2 cos x
2
′ 1 −(cos x)′ = sinx (sec x)′ = = 2 cos2 x cosx cos x
注：此法则可推广到多个中间变量的情形. ：此法则可推广到多个中间变量的情形. 例如,
y
dy dy d u dv = ⋅ ⋅ dx d u dv dx
u v x
= f ′(u) ⋅ϕ′(v) ⋅ψ′(x)
关键: 搞清复合函数结构, 由外向内逐层求导.
例4. 求下列各函数的导数. 求下列各函数的导数. 解:
= f ′( ln cos(ex ) )⋅ [−ex tan( ex )]
含义不同
f ′(u) u=lncos(ex )
例12. 设下列各函数的导数
(1) y = f [ f (sin x)] ; (2) y = f (ln x)e f (x) .
解：(1) y′ = f ′[ f (sin x)] ⋅ f ′(sin x) ⋅ cos x
1 f (x) (2) y′ = f ′(ln x) ⋅ ⋅e + f (ln x) ⋅ e f ( x) ⋅ f ′(x) (x x
四、小结
1. 基本初等函数的导数
(C)′ = 0 (sin x)′ = cos x (tanx)′ = sec2 x (secx)′ = sec xtan x (ax )′ = ax ln a
u ′ u′v − u v′ (3) ( ) = 2 v v u( x) 证: 设 f (x) = v( x) , 则有
u(x + h) u(x) − f (x + h) − f (x) v(x + h) v(x) f ′(x) = lim = lim h→0 h→0 h h
v(x + h) − v(x) u(x + h) − u(x) v(x) − u(x) h h
1 x x −1
2
2 sin x2arctan x2 −1 = 2xcos x e
+
e
sin x2
关键: 关键 看清函数结构
x +1 − x −1 求 ′ , y. 例7. y = x +1 + x −1
2x − 2 x −1 = x − x2 −1 解: Q y = 2 1 x ∴ y′ =1 − ⋅ (2x) =1− 2 x2 −1 x2 −1
第二节 函数的求导法则
一、四则运算求导法则 二、反函数的求导法则 三、复合函数求导法则 四、小结
第二章
已有的求导公式
(C )′ = 0
( sin x )′ = cos x
(cos x)′ = −sin x 1 ( ln x )′ = x
(x )′ = α x
α
α−1
一、四则运算求导法则
定理1. 定理 则 的和、 积、 (除分母 差、 商 为 0的点外) 都在点 x 可导, 且
2) 设 y = a (a > 0 , a ≠ 1) , 则 x = loga y , y ∈( 0 , + ∞)
x
1 = = (loga y)′
1
1 yln aຫໍສະໝຸດ = yln a特别当 a = e 时, ( ex )′ = ex 小结: 小结
( arcsin x)′ = ( arctan x)′ =
2
例8. 设 y = x
aa
+a
xa
+ a (a > 0),求 y′.
ax
解: y′ = a x
a aa −1
⋅ axa−1 + a ln a
xa
+ a ln a⋅ ax ln a
ax
例9. 求下列导数
解: (1) (x )′ = (e
µ
µ ln x
)′
= µ xµ−1
⋅ (µ ln x)′
⋅
x
µ
证: 在 x 处给增量 ∆x ≠ 0, 由反函数的单调性知
例3. 求反三角函数及指数函数的导数. 求反三角函数及指数函数的导数. 解: 1) 设 则
∴ cos y > 0 , 则
y ∈(− , ) , 2 2
1 1− sin y
2
π π
1 = = (sin y)′ cos y
利用 π arccos x = − arcsin x 2 类似可求得
∆f (x) = A ⇔ f (x)∆uA+α , 其 lim α = 0 y ∆u lim = f ′(u) +α = (∆x ≠ 0) 中 故有 x→x0 x→x0 ∆x ∆x ∆x dy ∆y = f ′(u)g′(x) ∴ = lim = lim d x ∆x→0 ∆x ∆x→0
x
′ = (ex ln x )′ (2) (x )
⋅ (xln x)′
= x (ln x +1)
x
例10. 设 解:
求
例11. 设 解:
求
1 = ⋅ (−sin(ex )) ⋅ ex cos(ex ) = −ex tan(ex )
若
f (lncos(ex )) 的导数. 存在 , 求
df = f ′( ln cos(ex ) ) ⋅ (ln cos(ex ))′ dx
3
解: y′ =(
x )′( x − 4cos x − sin1)
3
( x3 − 4cos x − sin1)′ + x
= 1 2 x ( x − 4cos x − sin1) + x ( 3 x2 + 4sin x)
3
1 y′ x=1 = (1− 4cos1− sin1) + (3+ 4sin1) 2 7 7 = + sin1− 2cos1 2 2
= sec x tan x
类似可证: (cot x)′ = −csc2 x , (csc x)′ = −csc xco t x .
二、反函数的求导法则
定理2. 定理 设 y = f (x)为x = f
−1
( y) 的 函 , f −1( y) 在 反 数
且 [ f −1( y)]′ ≠ 0,则有 y 的某邻域内单调可导,
(v(x) ≠ 0)
(1) (u ± v)′ = u′ ± v′ 证: 设 f (x) = u(x) ± v(x) , 则
f (x + h) − f (x) f ′(x) = lim h→0 h [ u(x + h) ± v(x + h) ] −[ u(x) ± v(x) ] = lim h→0 h u(x + h) − u(x) v(x + h) − v(x) = lim ± lim h→0 h→0 h h = u′(x) ± v′(x) 故结论成立.
= lim h→0 v(x + h)v(x) u′(xuv(xv(xu(x) v′(x) ± ) (x)) − ) = 故结论成立. 2 u(x + h)v(x) −u(v )(x) + h) x v(x C ′ − C v′ h 推论: ( ) ( 推论 v(x + h) v=x) 2 ( C为常数 ) v v
1
=
x + x2 +1 1
⋅ ( 1+
⋅ 2x 2 x +1
2
1
)
x2 +1
例5. 设
求
解：
例6. y = e 解:
sin x2
arctan x −1 , 求
2
y′ .
y′ = (e
sin x2
⋅ cos x2 ⋅ 2x) arctan x2 −1 1 sin x2 1 ⋅ ⋅ 2x ) +e ( 2 x 2 x2 −1
1 f ′(x) = −1 [ f ( y)]′
y= f ( x)
∆y 1 ∆y = f (x + ∆x) − f (x) ≠ 0, ∴ = ∆x ∆x ∆y 且由反函数的连续性知 ∆x →0 时 有∆y →0, 因此 必 ∆y 1 f ′(x) = lim = lim = 1 −1 ∆x→0 ∆x ∆y→0 ∆x [ f ( y)]′ ∆y
1 (loga x)′ = xln a 1
(x )′ = µ x (cos x)′ = − sin x (cot x)′ = − csc2 x (cscx)′ = − csc xcot x (ex )′ = ex
1 (lnx)′ = x
µ
µ−1
(arcsinx)′ =
1− x 1 (arctanx)′ = 1+ x2
= u′(x)v(x) + u(x)v′(x)
故结论成立.
推论: 推论 1) (Cu )′ = Cu′ ( C为常数 )
2) ( uvw)′ = u′vw+ uv′w+ uvw′ ′ ln x = 1 3) ( loga x )′ = ln a xln a