高等數(shù)學(xué)D2_2函數(shù)的求導(dǎo)法則

1、第二節(jié)二、反函數(shù)的求導(dǎo)法則二、反函數(shù)的求導(dǎo)法則 三、復(fù)合函數(shù)求導(dǎo)法則三、復(fù)合函數(shù)求導(dǎo)法則 四、初等函數(shù)的求導(dǎo)問題四、初等函數(shù)的求導(dǎo)問題 一、四則運算求導(dǎo)法則一、四則運算求導(dǎo)法則 函數(shù)的求導(dǎo)法則 第二章 解決求導(dǎo)問題的思路解決求導(dǎo)問題的思路:xxfxxfxfx)()(lim)(0( 構(gòu)造性定義 )求導(dǎo)法則求導(dǎo)法則其他基本初等其他基本初等函數(shù)求導(dǎo)公式函數(shù)求導(dǎo)公式0 xcosx1 )(C )sin(x )ln(x證明中利用了兩個重要極限初等函數(shù)求導(dǎo)問題初等函數(shù)求導(dǎo)問題本節(jié)內(nèi)容一、四則運算求導(dǎo)法則一、四則運算求導(dǎo)法則 定理定理1.的和、 差、 積、 商 (除分母為 0的點外) 都在點 x 可導(dǎo), 且下

2、面分三部分加以證明,并同時給出相應(yīng)的推論和例題 .可導(dǎo)都在點及函數(shù)xxvvxuu)()()()(xvxu及)()( )()() 1 (xvxuxvxu)()()()( )()()2(xvxuxvxuxvxu)()()()()()()()3(2xvxvxuxvxuxvxu)0)(xv此法則可推廣到任意有限項的情形.證證: 設(shè) 則vuvu )() 1 (故結(jié)論成立.例如, )()()(xvxuxfhxfhxfxfh)()(lim)(0hxvxuhxvhxuh)()()()(lim0hxuhxuh)()(lim0hxvhxvh)()(lim0)()(xvxuwvuwvu)(2)vuvuvu )(證證

3、: 設(shè), )()()(xvxuxf則有hxfhxfxfh)()(lim)(0hxvxuhxvhxuh)()()()(lim0故結(jié)論成立.)()()()(xvxuxvxuhhxuh )(lim0)(xu)(hxvhxv)( )(xu)(hxv推論推論: )() 1uC )()2wvuuC wvuwvuwvu )log()3xaaxlnlnaxln1( C為常數(shù) )例例1. 解解:xsin41(21)1sin, )1sincos4(3xxxy.1xyy 及求 y)(xx)1sincos4(213xxx23( xx)1xy1cos4)1sin43( 1cos21sin2727)1sincos4(3x

4、x)1sincos4(3xx)()( lim0 xvhxvh)()()()()()(xvhxvhxvxuxvhxuh)()(xvxu(3)2vvuvuvu證證: 設(shè))(xf則有hxfhxfxfh)()(lim)(0hh lim0,)()(xvxu)()(hxvhxu)()(xvxuhhxu )( )(xu)(xvhhxv )( )(xu)(xv故結(jié)論成立.)()()()()(2xvxvxuxvxu推論推論:2vvCvC( C為常數(shù) ) )(cscxxsin1x2sin)(sinxx2sin例例2. 求證,sec)(tan2xx證證: .cotcsc)(cscxxxxxxcossin)(tan

5、x2cosxx cos)(sin)(cossinxx x2cosx2cosx2sinx2secxcosxxcotcsc類似可證:,csc)(cot2xx.tansec)(secxxx )( xf二、反函數(shù)的求導(dǎo)法則二、反函數(shù)的求導(dǎo)法則 定理定理2. y 的某鄰域內(nèi)單調(diào)可導(dǎo), ,)()(1的反函數(shù)為設(shè)yfxxfy在)(1yf0 )(1yf且 ddxy或yxdd1 )(1yf11例例3. 求反三角函數(shù)及指數(shù)函數(shù)的導(dǎo)數(shù).解解: 1) 設(shè),arcsin xy 則,sin yx , )2,2(y)(arcsinx)(sinyycos1y2sin11211x類似可求得?)(arccosx,11)(arct

6、an2xx211)arccot(xx211xxxarcsin2arccos利用0cosy, 則2) 設(shè), )1,0(aaayx則),0(,logyyxa)(xa)(log1ya 1ayln1aylnaaxlnxxe)e( )arcsin(x211x )arccos(x211x )arctan(x211x )cotarc(x211xaaaxxln)(xxe)e(特別當(dāng)ea時,小結(jié)小結(jié):在點 x 可導(dǎo),三、復(fù)合函數(shù)求導(dǎo)法則三、復(fù)合函數(shù)求導(dǎo)法則定理定理3.)(xgu )(ufy 在點)(xgu 可導(dǎo)復(fù)合函數(shù) fy )(xg且)()(ddxgufxy在點 x 可導(dǎo),例如,)(, )(, )(xvvuu

7、fyxydd)()()(xvufyuvxuyddvuddxvdd關(guān)鍵: 搞清復(fù)合函數(shù)結(jié)構(gòu), 由外向內(nèi)逐層求導(dǎo).推廣推廣：此法則可推廣到多個中間變量的情形.例例4. 求下列導(dǎo)數(shù):. )(sh)3(;)()2(;)() 1 (xxxx解解: (1)(e)(lnxxxlne)ln(xxx1x)(e)(lnxxxxxx lne)ln(xxxx)1ln(x(2)(3)2ee)(shxxx2 xexexch說明說明: 類似可得;sh)(chxxaxxalne)(thx)(xaxxxchshth2eeshxxx;ch12x.lnaax例例5. 設(shè), )cos(elnxy 求.ddxy解解:xydd)cos(

8、e1x)sin(e(xxe)tan(eexx思考思考: 若)(uf 存在 , 如何求)cos(e(lnxf的導(dǎo)數(shù)?xfdd)(f ) )cos(e(lnx)cos(eln)(xuuf這兩個記號含義不同)cos(elnx例例6. 設(shè), )1(ln2xxy.y求解解:112xxy11212xx2112x記, )1(lnarsh2xxx則 )(arsh x112x(反雙曲正弦)其他反雙曲函數(shù)的導(dǎo)數(shù)看參考書自推. 2eeshxxx的反函數(shù)雙曲正弦四、初等函數(shù)的求導(dǎo)問題四、初等函數(shù)的求導(dǎo)問題 1. 常數(shù)和基本初等函數(shù)的導(dǎo)數(shù) (P95) )(C0 )(x1x )(sin xxcos )(cosxxsin

9、)(tan xx2sec )(cot xx2csc )(secxxxtansec )(cscxxxcotcsc )(xaaaxln )(exxe )(log xaaxln1 )(lnxx1 )(arcsin x211x )(arccosx211x )(arctan x211x )cot(arcx211x2. 有限次四則運算的求導(dǎo)法則 )(vuvu )( uCuC )( vuvuvuvu2vvuvu( C為常數(shù) )0( v3. 復(fù)合函數(shù)求導(dǎo)法則)(, )(xuufyxydd)()(xuf4. 初等函數(shù)在定義區(qū)間內(nèi)可導(dǎo)初等函數(shù)在定義區(qū)間內(nèi)可導(dǎo), )(C0 )(sin xxcos )(ln xx1由

10、定義證 ,說明說明: 最基本的公式uyddxudd其他公式用求導(dǎo)法則推出.且導(dǎo)數(shù)仍為初等函數(shù)且導(dǎo)數(shù)仍為初等函數(shù)例例7. 求解解:,1111xxxxy.y21222xxy12xx1 y1212x)2( x112xx例例8. 設(shè)),0( aaaxyxaaaxa解解:1aaaxayaaaxln1axaaaxaln求.yaaxln先化簡后求導(dǎo)例例9. 求解解:,1arctane2sin2xyx.y1arctan)(2xy ) (e2sin x2sinex2cos xx221x1212xx2x21arctan2x2sinex2cos x2sinex112xx關(guān)鍵關(guān)鍵: 搞清復(fù)合函數(shù)結(jié)構(gòu) 由外向內(nèi)逐層求導(dǎo)

11、例例10. 設(shè)求,1111ln411arctan21222xxxy.y解解: y22)1(1121x21xx) 11ln() 11ln(22xx111412x21xx1112x21xx2121xx221x21x231)2(1xxx內(nèi)容小結(jié)內(nèi)容小結(jié)求導(dǎo)公式及求導(dǎo)法則 (見P95 P96)注意注意: 1),)(vuuvvuvu2) 搞清復(fù)合函數(shù)結(jié)構(gòu) , 由外向內(nèi)逐層求導(dǎo) .41143x1.xx1431x思考與練習(xí)思考與練習(xí)對嗎?2114341xx2. 設(shè), )()()(xaxxf其中)(x在ax 因)()()()(xaxxxf故)()(aafaxafxfafax)()(lim)(axxaxax)()(lim)(limxax)(a正確解法:)(af 時, 下列做法是否正確?在求處連續(xù),由于 f (a) = 0，故3. 求下列函數(shù)的導(dǎo)數(shù)解解: (1)1bxaby2xa1bbxba(2) y)(x(1);(2).bxaayyxbxbabalnxabbaln或xabyababxln4. 設(shè)),99()2)(1()(xxxxxf).0(f 求解解: 方法方法1 利用導(dǎo)數(shù)定義.0)0()(lim)0(0 xfxffx)99()2)(1(lim0 xxxx!99方法方法2 利用求導(dǎo)公式.)(xf)(xx )99()2)(1(xxx)99()2)(1(xxx!99)0(