《高等数学》课程教学资源（课件讲稿）第二章_2.2函数的求导法则

第二节西数的求导法则一、和、差、积、商的求导法则二、反函数的求导法则三、复合茜数的求导法则四、基本求导法则与求导公式五、小结思考题

第二节 函数的求导法则 • 一、和、差、积、商的求导法则 • 二、反函数的求导法则 • 三、复合函数的求导法则 • 四、基本求导法则与求导公式 • 五、小结 思考题

一、和、差、积、商的求导法则定理1 如果函数u(x)，v(x)在点x处可导,则它们的和、差、积、商(分母不为零)在点x处也可导，并且(1) [u(x)±v(x))' = u'(x)±v'(x);(2) [u(x) · v(x))' = u(x)v(x) +u(x)v'(x);u'(x)v(x)-u(x)v(x)X(3) [(v(x) ± 0)v(x)v(x)

一、和、差、积、商的求导法则 (2) [ ( ) ( )] ( ) ( ) ( ) ( ); u x v x u x v x u x v x  = +    定理1 ( ), ( ) , ( ) , u x v x x x 如果函数 在点 处可导 则它 们的和、差、积、商 分母不为零 在点 处也 可导 并且 (1) [ ( ) ( )] ( ) ( ); u x v x u x v x  =     2 ( ) ( ) ( ) ( ) ( ) (3) [ ] ( ( ) 0). ( ) ( ) u x u x v x u x v x v x v x v x   −  = 

证: (2)[u(x).v(x)' = u'(x)v(x)+u(x)v'(x);[u(x)(x)]u(x + Ar)v(x + Ax)-u(x)v(x)lim=ArAr→0u(x+△x)-u(x)v(x+Ar)-v(x)v(x + Ax)+ u(x)limAxArAr-→0u(x+△x)-1v(x + Ar) -v(x)u(xlim v(x + Ax)+u(x) lim= limAxArAr-→0Ar-→0Ar-→0=u(x)v(x)+u(x)v'(x)lim v(x+Ax)=v(x) 是由于 v'(x) 的存在其中Ar-→0

证： (2) [ ( ) ( )] ( ) ( ) ( ) ( ); u x v x u x v x u x v x  = +    ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 ( ) ( ) ( ) ( ) lim ( ) ( ) ( ) ( ) lim ( ) ( ) ( ) ( ) ( ) ( ) lim lim ( ) ( ) lim x x x x x u x v x u x x v x x u x v x x u x x u x v x x v x v x x u x x x u x x u x v x x v x v x x u x x x u x v x u x v x  →  →  →  →  →     +  +  − =    +  − +  − = +  +       +  − +  − = +  +   = +   其中 ( ) 0 lim ( ) x v x x v x  → +  = 是由于 v x ( ) 的存在

u(x)设 f(x)=证(3)(v(x) ± 0),v(x)f(x+h)- f(x)f'(x)= limhh-→0u(x + h)u(x)v(x)v(x + h)= limhh-→0u(x+h)v(x)-u(x)v(x+h= limh-→0v(x +h)v(x)h

证(3) , ( ( ) 0), ( ) ( ) ( ) = v x  v x u x 设 f x h f x h f x f x h ( ) ( ) ( ) lim 0 + −  = → v x h v x h u x h v x u x v x h h ( ) ( ) ( ) ( ) ( ) ( ) lim 0 + + − + = → h v x u x v x h u x h h ( ) ( ) ( ) ( ) lim 0 − + = →

[u(x + h)-u(x)lv(x)-u(x)[v(x + h)-v(x)= limh-→0v(x +h)v(x)hu(x+ h)-u(x)v(x +h) -v(x)2. (x) -u(x).hh= limh-→0v(x + h)v(x)u'(x)v(x)-u(x)v(x)[v(x)]}?:f(x)在x处可导

v x h v x h u x h u x v x u x v x h v x h ( ) ( ) [ ( ) ( )] ( ) ( )[ ( ) ( )] lim 0 + + − − + − = → ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) lim 0 v x h v x h v x h v x v x u x h u x h u x h + + −  −  + − = → 2 [ ( )] ( ) ( ) ( ) ( ) v x u x v x − u x v x = f (x)在x处可导

推广(Ztu,(x)l'=Eu(x);(1)i=1i=1(2)[Cu(x)}' = Cu'(x);1(3) [/Iu,;(x)' = u'(x)u,(x)...u.,(x)i=1+...+u,(x)u,(x)...u,(x)u,(x)u (x);i=1 k=1kti

推广 1 1 (1) [ ( )] ( ); n n i i i i u x u x = =     = (2) [ ( )] ( ); Cu x Cu x   = 1 2 1 1 2 1 1 (3) [ ( )] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ); n i n i n n n i k i k k i u x u x u x u x u x u x u x u x u x = = =   = + +  =    

例1 求y=x3-2x2+sinx的导数解 y'= 3x2-4x+ cos x.例2 求y= sin2x·lnx的导数解y=2sinx·cosx·ln xy' = 2cos x .cosx.ln x+ 2sin x(-sin x).ln x+ 2sin x.cosx.x1=2cos2xlnx+=sin2x.x

例 1 2 sin . 求 y = x3 − x2 + x 的导数 解 2 y x x x  = − + 3 4 cos . 例2 求 y = sin2x  ln x 的导数 . 解  y = 2sin x  cos x  ln x y = 2cos x  cos x  ln x+ 2sin x (− sin x) ln x x x x 1 + 2sin  cos  sin 2 . 1 2cos 2 ln x x = x x +

例3求y=tanx的导数解 y'=(tanx)'=(sinz)cosx(sin x)'cos x - sin x(cos x)cos" x1cos x + sin' x=sec"x2cos"xcos"x即(tan x) = sec’ x.同理可得(cot x)' = -csc2 x

例3 求 y = tan x 的导数 . 解 ) cos sin  = (tan ) = (  x x y x x x x x x 2 cos (sin ) cos − sin (cos ) = x x x 2 2 2 cos cos + sin = x x 2 2 sec cos 1 = = 2 即 (tan ) sec . x x  = 2 同理可得 (cot ) csc . x x  = −

例4 求 y= secx的导数,解'=(secx)=(cosx_ -(cosx)cosxsinx= secx tanx.2cosx同理可得(cscx)' = -csc xcot x

例4 求 y = sec x 的导数 . 解 ) cos 1  = (sec ) = (  x y x x x 2 cos − (cos ) = = sec x tan x. x x 2 cos sin = 同理可得 (csc ) csc cot . x x x  = −

例5. y=/x(x3 -4cosx-sin1),求y'及j'|x=1 解: y' =(Vx)(x3 -4cosx -sin1)+/x(x3 - 4cos x -sin1)1(x3 -4cosx -sin1) + /x (3x + 4sin x)2Vx1(1-4cos1- sin1)+(3+4sin1)x=127.7sin1-2cos1+一一22

解: 3 1 5. ( 4cos sin1) , . x y x x x y y = 例 = − − 求   及 3 y x x x   = − − ( ) ( 4cos sin1) 3 + − − x x x ( 4cos sin1) 1 3 ( 4cos sin1) 2 x x x = − − 1 1 (1 4cos1 sin1) ( 3 4sin1) 2 x y =  = − − + + 7 7 sin1 2cos1 2 2 = + − 2 + + x x x ( 3 4sin )