1、資中縣龍結(jié)中學(xué)數(shù)學(xué)組資中縣龍結(jié)中學(xué)數(shù)學(xué)組(第二課時(shí)第二課時(shí))資中縣龍結(jié)中學(xué)數(shù)學(xué)組資中縣龍結(jié)中學(xué)數(shù)學(xué)組一、基本初等函數(shù)的導(dǎo)數(shù)公式表一、基本初等函數(shù)的導(dǎo)數(shù)公式表為 常 數(shù)則則則則則則則則x xx xa a若若 f f ( ( x x ) ) = = c c ( ( c c) ) , ,f f ( ( x x ) ) = =若若 f f ( ( x x ) ) = = x x ( (Q Q ) ) , ,f f ( ( x x ) ) = =若若 f f ( ( x x ) ) = = s s i i n n x x , ,f f ( ( x x ) ) = =若若 f f ( ( x x ) )

2、= = c c o o s s x x , ,f f ( ( x x ) ) = =若若 f f ( ( x x ) ) = = a a , ,f f ( ( x x ) ) = =若若 f f ( ( x x ) ) = = e e , ,f f ( ( x x ) ) = =若若 f f ( ( x x ) ) = = l l o o g g x x , ,f f ( ( x x ) ) = =若若 f f ( ( x x1 1 . .2 2 . .3 3 . .4 4 . .5 5 . .6 6 . .) ) = = l l n n x x , ,f f7 7 . .8 8 . .( (

3、 x x ) ) = =0 0 - -1 1 x xcosxcosx-sinx-sinxlnax xa aexex1 1x xl ln na a1 1x x資中縣龍結(jié)中學(xué)數(shù)學(xué)組資中縣龍結(jié)中學(xué)數(shù)學(xué)組二、導(dǎo)數(shù)的運(yùn)算法則二、導(dǎo)數(shù)的運(yùn)算法則f f( (x x) ) 1 1g g( (x x) ). .= = f f( (x x) )2 2g g( (x x) ). .= =f f( (x x) ) = = ( (g g g g( (x x) )3 3. .( (x x) )0 0) )別為數(shù)c cg g( (x x) ) = = 特特地地( (c c 常常 ) )f f( (x x) )g g( (x

4、 x) )f f( (x x) )g g( (x x) )+ +f f( (x x) )g g( (x x) )c cg g ( (x x) )2 2f f( (x x) )g g( (x x) )- -f f( (x x) )g g( (x x) )g g( (x x) )資中縣龍結(jié)中學(xué)數(shù)學(xué)組資中縣龍結(jié)中學(xué)數(shù)學(xué)組數(shù)數(shù)導(dǎo)導(dǎo)數(shù)數(shù)嗎嗎2 22 2你你能能求求下下列列函函的的？( (1 1) )y y= =( (3 3x x+ +1 1) ) ( (2 2) )y y= =( (- -3 3x x+ +1 1) ) 2 2( (1 1) )y y = =( (9 9x x + +6 6x x+ +1

5、 1) )= =18x+618x+6 2 2( (2 2) )y y = =( (9 9x x - -6 6x x+ +1 1) )= =18x-618x-6你還有其它方法嗎？你還有其它方法嗎？ y y = =1 18 8x x+ +6 6= =2 2( (3 3x x+ +1 1) ) 3 3= =(1)(1)令令u=3x+1u=3x+1，2 2( (u u ) )u u(2)(2)令令u=-3x+1u=-3x+1，2 2y y = =( (u u ) )u u = =2 2u u ( (- -3 3) )= = - -6 6u u = =18x-618x-6資中縣龍結(jié)中學(xué)數(shù)學(xué)組資中縣龍結(jié)中

6、學(xué)數(shù)學(xué)組 對(duì)于函數(shù)對(duì)于函數(shù)y= f (x),令令u= (x),若若y=f(u)是是中間變量中間變量u的函數(shù)的函數(shù), u= (x)是自變量是自變量x的函數(shù)的函數(shù),則則稱稱y= f (x)是自變量是自變量x的復(fù)合函數(shù)的復(fù)合函數(shù). 設(shè)函數(shù)設(shè)函數(shù) 在點(diǎn)在點(diǎn)x處有導(dǎo)數(shù)處有導(dǎo)數(shù) ,函數(shù)函數(shù)y=f(u)在點(diǎn)在點(diǎn)x的對(duì)應(yīng)點(diǎn)的對(duì)應(yīng)點(diǎn)u處有導(dǎo)數(shù)處有導(dǎo)數(shù) ,則復(fù)合則復(fù)合函數(shù)函數(shù) 在點(diǎn)在點(diǎn)x處也有導(dǎo)數(shù)處也有導(dǎo)數(shù),且且 或記或記)(xu )(xux )(ufyu ;xuxuyy ).()()(xufxfx )(xfy 求復(fù)合函數(shù)的導(dǎo)數(shù)求復(fù)合函數(shù)的導(dǎo)數(shù),關(guān)鍵在于分清函數(shù)的復(fù)合關(guān)鍵在于分清函數(shù)的復(fù)合關(guān)系關(guān)系,合理選定中間變

7、量合理選定中間變量,明確求導(dǎo)過程中每次是明確求導(dǎo)過程中每次是哪個(gè)變量對(duì)哪個(gè)變量求導(dǎo)哪個(gè)變量對(duì)哪個(gè)變量求導(dǎo),一般地一般地,如果所設(shè)中間如果所設(shè)中間變量可直接求導(dǎo)變量可直接求導(dǎo),就不必再選中間變量就不必再選中間變量.資中縣龍結(jié)中學(xué)數(shù)學(xué)組資中縣龍結(jié)中學(xué)數(shù)學(xué)組解解:(1):(1)(2)(2)(3)(3)(4)(4)1數(shù)導(dǎo)數(shù)2 2- -0 0. .0 05 5x x+ +1 1求求下下列列函函的的 ( (1 1) )y y = = 2 2x x+ +3 3( (2 2) )y y = =l ln n( (例例 3 3x x+ +2 2) ) ( (3 3) )y y = = e e( (4 4) )y

8、y = =s si in n x x+ +令令u=2x+3u=2x+3，那，那么么2 2y y = =( (u u ) )u u = =4u=4u= 8x+128x+12令令u=3x+2u=3x+2，那，那么么y y = =( (l ln nu u) )u u = =3 3u u3 33 3x x+ +2 2令令u=-0.05x+1u=-0.05x+1，那，那么么uy y = =( (e e ) )u uu= =- -0 0. .0 05 5e e = =- -0 0. .0 05 5x x+ +1 1- -0 0. .0 05 5e e則令令u u = =x x+ + , ,y y = =(

9、 (s si in nu u) ) u u = =cosu uc co os s( (x x+ += =) )資中縣龍結(jié)中學(xué)數(shù)學(xué)組資中縣龍結(jié)中學(xué)數(shù)學(xué)組例例2 2 求證求證: :雙曲線雙曲線C1:x2-y2=5C1:x2-y2=5與橢圓與橢圓C2:4x2+9y2=72C2:4x2+9y2=72在交在交點(diǎn)點(diǎn) 處的切線互相垂直處的切線互相垂直. .證明證明: :由于曲線的圖形關(guān)于坐標(biāo)軸對(duì)稱由于曲線的圖形關(guān)于坐標(biāo)軸對(duì)稱, ,故只需證明其中一故只需證明其中一 個(gè)交點(diǎn)處的切線互相垂直即可個(gè)交點(diǎn)處的切線互相垂直即可. .聯(lián)立兩曲線方程解得第一象限的交點(diǎn)為聯(lián)立兩曲線方程解得第一象限的交點(diǎn)為P(3,2),P(3

10、,2),不妨不妨證明過證明過P P點(diǎn)的兩條切線互相垂直點(diǎn)的兩條切線互相垂直. .由于點(diǎn)由于點(diǎn)P P在第一象限在第一象限, ,由由x2-y2=5x2-y2=5得得;23|31 xyk同理由同理由4x2+9y2=724x2+9y2=72得得.32|32 xyk因?yàn)橐驗(yàn)閗1k2=-1,k1k2=-1,所以兩條切線互相垂直所以兩條切線互相垂直, ,從而命題成立從而命題成立. .,2 2y y= = x x - -5 52 2x xy y = =x x - -5 5,2 24 4y =8 -xy =8 -x9 992 2- -4 4x x y y = =4 48 8 - -x x9 9資中縣龍結(jié)中學(xué)數(shù)學(xué)

11、組資中縣龍結(jié)中學(xué)數(shù)學(xué)組 對(duì)于函數(shù)對(duì)于函數(shù)y= f (x),令令u= (x),若若y=f(u)是是中間變量中間變量u的函數(shù)的函數(shù), u= (x)是自變量是自變量x的函數(shù)的函數(shù),則則稱稱y= f (x)是自變量是自變量x的復(fù)合函數(shù)的復(fù)合函數(shù). 設(shè)函數(shù)設(shè)函數(shù) 在點(diǎn)在點(diǎn)x處有導(dǎo)數(shù)處有導(dǎo)數(shù) ,函數(shù)函數(shù)y=f(u)在點(diǎn)在點(diǎn)x的對(duì)應(yīng)點(diǎn)的對(duì)應(yīng)點(diǎn)u處有導(dǎo)數(shù)處有導(dǎo)數(shù) ,則復(fù)合則復(fù)合函數(shù)函數(shù) 在點(diǎn)在點(diǎn)x處也有導(dǎo)數(shù)處也有導(dǎo)數(shù),且且 或記或記)(xu )(xux )(ufyu ;xuxuyy ).()()(xufxfx )(xfy 求復(fù)合函數(shù)的導(dǎo)數(shù)求復(fù)合函數(shù)的導(dǎo)數(shù),關(guān)鍵在于分清函數(shù)的復(fù)合關(guān)鍵在于分清函數(shù)的復(fù)合關(guān)系關(guān)系,合理選定中間變量合理選定中間變量,明確求導(dǎo)過程中每次是明確求導(dǎo)過程中每次是哪個(gè)變量對(duì)哪個(gè)變量求導(dǎo)哪個(gè)變量對(duì)哪個(gè)變量求導(dǎo),一般地一般地,如果所設(shè)中間如果所設(shè)中間變量可