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