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第二节求导法则与导数公式导数的四则运算 基本初等函数的导数 复合函数的导数 反函数求导法则导数的四则运算（1）设 u ( x) v( x) 在x可导，则[u ( x) ± v( x)]′ = u ′( x) ± v′( x) 设 y = g ( x) = u ( x) + v( x)Δy = g ( x + Δx) − g ( x) = [u ( x + Δx) + v( x + Δx)] − [u ( x) + v( x)]= [u ( x + Δx) − u ( x)] + [v( x + Δx) − v( x)]= Δu + Δv Δy Δu Δv lim Δy = lim [ Δu + Δv ] = u ′( x) + v′( x) + = Δx Δx Δx Δx →0 Δx Δx →0 Δx Δx推广[u1 ( x) ± u2 ( x) ±′ ( x) ± u2 ′ ( x) ± u n ( x)]′ = u1′ ( x). ± un[u ( x) ± v( x)]′ = u ′( x) ± v′( x)[u1 ( x) ± u2 ( x) ±例 解′ ( x) ± u2 ′ ( x) ± u n ( x)]′ = u1′ ( x). ± unf ( x) = x + sin x − cos x + 9 求其导数 f ′( x) = ( x + sin x − cos x + 9)′ = ( x )′ + (sin x)′ − (cos x)′ + (9)′= 1 / 2 x + cos x + sin x(2)设u ( x) , v( x)在x可导,则[u ( x)v( x)]′ = u ( x)v′( x) + u ′( x)v ( x ) 设 y = g ( x ) = u ( x )v ( x )Δy = g ( x + Δx ) − g ( x ) = u ( x + Δ x ) v ( x + Δx ) − u ( x ) v ( x ) = u ( x + Δ x ) v ( x + Δ x ) − u ( x ) v ( x + Δx ) + u ( x ) v ( x + Δ x ) − u ( x ) v ( x )= Δu ⋅ v( x + Δx) + u ( x)Δv. Δv Δy Δu = v ( x + Δx ) + u ( x ) . Δx Δx Δx Δy Δu Δv lim = lim ⋅ lim v( x + Δx) + u ( x) ⋅ lim Δx → 0 Δx Δx → 0 Δx Δx → 0 Δx → 0 Δx= u ( x)v′( x) + u ′( x)v( x).[u ( x)v( x)]′ = u ( x)v′( x) + u ′( x)v ( x )[cu ( x)]′ = cu ′( x) （常数因子可以提出来） 特别：例、求 f (x) = 7 x cosx 的导数 解 f ′( x) = (7 x cos x)′ = 7[( x ) cos x +′′ x (cos x ) ]cos x = 7[ − x sin x] 2 x推广 (u ( x)v( x) w( x))′轮流求导= u ′( x)v( x) w( x) + u ( x)v′( x) w( x) + u ( x)v( x) w′( x)[u1 ( x)u2 ( x)′ ( x)u2 ( x) un ( x)]′ = u1 ′ ( x) + u1 ( x)u 2 un ( x) +un ( x) + u1 ( x)u2 ( x) ′ ( x). un例、求 f ( x ) = 4 x 2 ⋅ ln x ⋅ cos x 的导数 解 f ′(x) = (4x2 ⋅ ln x ⋅ cosx)′ = 4(x2 ⋅ ln x ⋅ cosx)′1 = 4(2x ⋅ ln x ⋅ cosx + x2 ⋅ ⋅ cosx − x2 ⋅ ln x ⋅ sin x) x = 4(2x ⋅ ln x ⋅ cosx + x cosx − x2 ⋅ ln x ⋅ sin x)（3）设′ u ( x) ⎡ u ( x) ⎤ u ′( x)v( x) − u ( x)v′( x) 设 y = g ( x) = . ⎢ v( x) ⎥ = v( x) 2 [v( x)] ⎣ ⎦ Δy = g ( x + Δx ) − g ( x ) u ( x + Δx) u ( x) u ( x + Δx)v( x) − u ( x)v( x + Δx) = − = v( x + Δx) v( x) v( x + Δx)v( x) u ( x + Δx)v( x) − u ( x)v( x) + u ( x)v( x) − u ( x)v( x + Δx) = v( x + Δx)v( x) Δuv( x) − u ( x)Δv = v( x + Δx)v( x) Δu Δv ⋅ v( x) − u ( x) ⋅ Δy Δx Δx = 因为u,v可导，所以也连续 Δx v( x + Δx)v( x)u ( x) , v( x) 在x可导 v( x ) ≠ 0u ′( x) ⋅ v( x) − u ( x) ⋅ v′( x) Δy lim = Δx →0 Δx [v( x)]2例、求y=tanx的导数 sin x ∵ y = tan( x) = cos x ′ sin x ⎞ (sin x)′ cos x − sin x(cos x)′ cos 2 x + sin 2 x ⎛ ∴ y′ = ⎜ ⎟= = 2 2 x cos cos x cos x ⎠ ⎝(tan x)′ = sec 2 x(cot x)′ = − csc 2 x= sec 2 x′ 1 ⎞ − v′( x) 特别地 ⎛ ⎜ ⎜ v( x) ⎟ ⎟ = [v( x)]2 ⎝ ⎠ ′ sin x 1 ⎞ ⎛ = tan x ⋅ sec x (sec x)′ = ⎜ ⎟ = 2 ⎝ cos x ⎠ cos x (csc x)′ = − cot x ⋅ csc x例x2 y= x 2(u ± v )′ = u ′ ± v′ (uv )′ = u ′v + uv′′ u ⎛ ⎞ u ′v − uv′ ⎜ ⎟ = v2 ⎝v⎠解： y′ = ( x 2 ⋅ 2− x )′= 2 x 2− x + x 2 (−2− x ln 2) 2 x − x 2 ln 2 = . x 2基本的初等函数的求导公式c′ = 0(c为常数 ).( x a )' = ax a −1 (a为实数 ) .y′ y == 2x 例： ，求 y′ 1 x x x − 1 2 y′ = x 2 ⎛ ⎞′ 7 ′ − 1 ⎛ ⎞ 1 −2 ⎜ ⎟ 8 y′ = − x = − 解： x = ⎜ ⎟ 2⎟ ⎜ x ⎠ ⎝ x 3x x ⎠ ⎝ 1 −2 1 y′ = − x 15 =− 2 7 − 8 2 x3 =− x . 81y=x ,2y= x =x 1 y = = x −1 x 1 − 1 y= =x 2 x1 2基本的初等函数的求导公式c′ = 0(c为常数 ).(a x )' = a x ⋅ ln a (a > 0, a ≠ 1). 1 1 (log a x)' = ⋅ (a > 0, a ≠ 1). x ln a ( x )′ = 3 x3 2( x a )' = ax a −1 (a为实数 ) . (e x )' = e x . 1 (lnx)' = . x(3x )' = 3x ln 3 (π x )′ = π x ln π ((tan α ) x )′ = (tan α ) x ⋅ ln tan α( xπ )′ = π xπ −1 ( x tan α )′ = tan α ⋅ x tan α −1识别函数关键常数、变量所在位置幂函数 例如 指数函数ax＝aa＝xx＝xa识别对数函数log a x＝ log x a ＝log x x基本的初等函数的求导公式c′ = 0(c为常数 ).( x a )' = ax a −1 (a为实数 ) .(e x )' = e x . 1 1 1 (log a x)' = ⋅ (a > 0, a ≠ 1). (lnx)' = . x ln a x (sin x)' = cos x. 比较两边 (cos x)' = − sin x. (tan x)' = sec 2 x. (sec x)' = sec x ⋅ tan x. 1 (arcsin x)' = . 2 1− x 1 (arctan x)' = . 2 1+ x (cot x)' = − csc 2 x. (csc x)' = − csc x ⋅ cot x. 1 (arccos x)' = − . 1 − x2 1 . (arc cot x)' = − 2 1+ x(a x )' = a x ⋅ ln a (a > 0, a ≠ 1).例．设，求f ′ (1) , f ′( ) 8 4 π 解： f ′( x ) = ( x sin x)′ + (tan )′ π π 8 f ′( )＝{ f ( )}′ ′ 4 4 ′ = x sin x + x (sin x ) f ′(1)＝{ f (1)}′ 1 sin x + x cos x = 2 x 1 f ′(1) = sin 1 + cos1 2 π 1 π π π 3π f ′( ) = sin + cos = 4 4 2 4 4 π π 注： tan 是常数，其导数等于零； 8f ( x ) = x sin x + tanππ( )x +2 x− π , 求y'. 例 设y = x解1 1 x − πx + 2 − y' = ( )' = ( x 2 − π + 2 x 2 )' x=1 ( x 2 )'− 1 2− ( π )' + 2( x1 + 2 ⋅ (− ) x 2 − 1 x x .−3 2−1 2 )'求导前先化简 可减少计算量1 = x 2 =1 2 x1 例. 求 y = 的导数 1+ x1 ′(1 + x ) − (1 + x )′ ( 1 ) 解： y′ = ( )′ = 1+ x (1 + x ) 21 2 x =− = 2 2 x (1 + x ) 2 (1 + x )1 − f ′( x) )′ = 2 一般 ( , 其中f (x)可导, f (x) ≠ 0 f ( x) f ( x)−1复合函数的导数若函数 u = g ( x ) 在x可导, 函数 y = f (u ) 在u可导 则复合函数 y= f [ g ( x )] 在x 可导 且{ f [ g ( x)]}′ = f ′(u ) g ′( x)Δy = f ( g ( x + Δx)) − f ( g ( x)) Δu = g ( x + Δx ) − g ( x ) ,= f (u + Δu ) − f (u )Δy Δy Δu = ⋅ Δx Δu Δxlim Δu = 0 所以 Δ x →0( Δu ≠ 0)Δy Δu Δy lim = lim ⋅ lim Δx →0 Δx Δx →0 Δu Δx →0 Δx因为u在 x 可导，所以必连续Δy Δu = lim ⋅ lim Δu → 0 Δu Δx → 0 Δx分析{ f [ g ( x)]}′ = f ′(u ) g ′( x){(6 x + 7) 2 }′ = 2(6 x + 7) ⋅ (6 x + 7)′y = u2 y′ = (u 2 )′ ⋅ (6 x + 7)′ = 12uu=6x+7= 12(6 x + 7)设 y = f (u ), u = ϕ ( x) , 则复合函数 y = f [ϕ ( x)] 的导数为dy dy du = dx du dx或{ f [ϕ ( x )]}′ = f ′(u )ϕ ′( x )例.求y = sin2x的导数 解：y = sin2x是由y = sinu，u = 2x复合而成dy du y′ = ⋅ = cos u ⋅ 2 = 2cos 2 x du dx例 设 y=sin3 x，求 y'. 解 令y = u 3，u = sin x，则dy dy du = ⋅ dx du dx = 3u 2 cos x = 3 sin 2 x ⋅ cos x.例. 求y = (3x2+1)100的导数 解： y = u100，而 u = 3x2+1 由公式dy du y′ = ⋅ du dx= 100 u ⋅ 6 x99= 600x(3x 2 + 1)99)2ctg ( )4(′=′xy )2ctg (2ctg21′⋅=x x)2()2csc (2ctg 212′⋅−⋅=x x x )21(2ctg22csc 2⋅−=x x 2csc 2412x x tg ⋅−=})]([{′x f ϕ)()]([u f x f ′=′ϕ表示复合函数对自变量x 求导；而对中间变量求导。
