天津工业大学数学系：《高等数学》课程电子教案（PPT课件）第二章 导数与微分（2.2）函数的求导法则

⑩天掌 Teaching Plan on Advanced Mathematics 第二节函数的求导法则 函数的和、差、积、商的求导法则 反函数的求导法则 、复合函数的求导法则 四、基本求导法则与导数公式 「返回 Tianjin Polytechnic Moiwendity w

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 第二节 函数的求导法则 一、函数的和、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、基本求导法则与导数公式 返回

⑩天掌 Teaching Plan on Advanced Mathematics 函数的和、差、积、商的求导法则 定理如果函数u(x),v(x)在点x处可导则它 们的和、差、积、商分母不为零)在点处也 可导,并且 (1)[(x)±v(x)=u(x)±v(x); (2)[u(x)·v(x)=u'(x)v(x)+u(x)v(x) u(xy u(xv(x)-u(x)v(r) (3)出 (v(x)≠0) vx v(x

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 一、函数的和、差、积、商的求导法则 定理 可导,并 且 们的和、差、积、商(分母不为零)在点 处 也 如果函数 在 点 处可导,则 它 x u(x), v(x) x ( ( ) 0). ( ) ( ) ( ) ( ) ( ) ] ( ) ( ) (3)[ (2)[ ( ) ( )] ( ) ( ) ( ) ( ); (1)[ ( ) ( )] ( ) ( ); 2   −   =   =  +    =    v x v x u x v x u x v x v x u x u x v x u x v x u x v x u x v x u x v x

⑩天掌 Teaching Plan on Advanced Mathematics 证 x)士以(x 加nu(x+△x)±v(x+Ax)-以(x)±v(x) △r→>0 △ n(x+△x)-(x) v(x+△x)-v(x) ±im △→>0 △ △->0 △ 于是法则(1)获得证明.法则(1)可简单地表示为 (uv)=u'±p

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 证 x v x x v x x u x x u x x u x x v x x u x v x u x v x x x x  +  −   +  − =  +   +  −  =    →  →  → ( ) ( ) lim ( ) ( ) lim [ ( ) ( )] [ ( ) ( )] lim [ ( ) ( )] 0 0 0 于是法则(1)获得证明. 法则(1)可简单地表示为 (u  v) = u   v 

⑩天掌 Teaching Plan on Advanced Mathematics 证(2) Lu(xv(xr= lim u(x+△x)νv(x+△x)-u(x)v(x) At→>0 △r im/4(x+△x)-l(x) v(x+△x)+u(x) v(x+△x)-v(x) Ar->0 △ nu(x+△x)-u(x) imv(x+△x) Ar→>0 △v Ar→>0 tu(x)lim v(x+△x)-v(x) △x→0 △ u(rv(r)+u(xv(x) 于是法则(2)获得证明.法则(2)可简单地表示为 (uv)=uv±up

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics ( ) ( ) ( ) ( ) ] ( ) ( ) ( ) lim lim ( ) ( ) ( ) lim ] ( ) ( ) ( ) ( ) ( ) ( ) lim[ [ ( ) ( )] ( ) ( ) [ ( ) ( )] lim 0 0 0 0 0 u x v x u x v x x v x x v x u x v x x x u x x u x x v x x v x v x x u x x u x x u x x u x x v x x u x v x u x v x x x x x x =  +   +  − + +   +  − =  +  − +  +  +  − =  +  +  −  =  →  →  →  →  → 证(2) 于是法则(2)获得证明. 法则(2)可简单地表示为 (uv) = u  v  uv 

⑩天掌 Teaching Plan on Advanced Mathematics 证(3) u(x+△x)a(x) [Wr=lim v(x+Ax)v(r) Ax→0 △ = limr u(x+△x)v(x)-a(x)w(x+△u x→》0 W(x+△x)w(x)△x =lma(x+△)-a(x)p(x)-a(x)v(x+△x)-(x Ar→>0 v(x+△x)v(x)△x u(x+△x)-u(x) v(x+△x)-ν(x) v(r-u(r u(x) lim △r (x+△x)w(x) u(r)v(r-u(x)v(x)

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics ( ) ( ) ( ) ( ) ( ) ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) lim ( ) ( ) [ ( ) ( )] ( ) ( )[ ( ) ( )] lim ] ( ) ( ) ( ) ( ) ( ) ( ) lim[ ( ) ( ) ( ) ( ) ] lim ( ) ( ) [ 2 0 0 0 0 v x u x v x u x v x v x x v x x v x x v x v x u x x u x x u x u x v x x v x x u x x u x v x u x v x x v x v x x v x x u x x v x u x v x x x v x u x v x x u x x v x u x x x x x  −  = +   +  − −  +  − = +   +  − − +  − = +   +  − +  =  − +  +   =  →  →  →  → 证(3)

⑩天掌 Teaching Plan on Advanced Mathematics 于是法则(3)获得证明.法则(3)可简单地表示为 在法则(2)中,当vx)=c(C为常数)时,有

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 于是法则(3)获得证明. 法则(3)可简单地表示为 . 2 v u v uv v u  −  =        在法则(2)中,当 v(x) = c (C 为常数)时,有 (Cu) = Cu

⑩天掌 Teaching Plan on Advanced Mathematics 例1求y=3x3-2x2+2sinx-9的导数 解y=9x2-4x+2cos(x) 例2f(x)=x3+4csx-simn,求f(x)及」 解∫(x)=3x2-4sinx 丌、3 ∫()=,x2-4 24

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例1 3 2 2sin 9 . 求 y = x 3 − x 2 + x − 的导数 解 9 4 2cos( ) 2 y  = x − x + x 例2 , 求 及 2 ( ) 4cos sin 3 x f x = x + x − ) 2 ( ) (  f  x f 解 4 4 3 ) 2 ( ( ) 3 4sin 2 2  = −  = −   f f x x x

⑩天掌 Teaching Plan on Advanced Mathematics 例3y= e(sinx+cosx,求y A y=(e)(sinx+cos x)+e(sin x+cos x) (e )sin x+cos x)+e(cos x-sin x) 2e cos x

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例3 y e (sin x cos x), 求 x = + y  解 e x e x x e x x y e x x e x x x x x x x 2 cos ( )(sin cos ) (cos sin ) ( ) (sin cos ) (sin cos ) = = + + −  =  + + + 

⑩天掌 Teaching Plan on Advanced Mathematics 例4求y=tanx的导数 SInd 解y=(tanx)y= cos X (sin x) cosx-sin x(cos x) cos x cosx+sin x 1 sec r cos cos x anr)=sec x 同理可得(cotx)=-csc2x

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例4 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 = = (tan ) sec . 2 即 x  = x (cot ) csc . 2 同理可得 x  = − x

⑩天掌 Teaching Plan on Advanced Mathematics 例5求y=sex的导数. 解y=(secx)y=( cosI (1)cos x-1(cos x) cos sin(x) =secxtan x cos x 即( (secx)= secx tan x 同理可得(cscx)=- cscxcot x 「返回 Tianjin Polytechnic Moiwendity w

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例5 求 y = secx的导数. 解 ) cos 1  = (sec ) = (  x y x x x x 2 cos (1)cos −1(cos ) = x x 2 cos sin( ) = = secx tan x 即 (secx) = secxtan x. 同理可得 (csc x) = −csc xcot x. 返回