高數(shù)二章課件04隱函數(shù)參數(shù)導(dǎo)數(shù).pdf

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1、2.4由方程所確定的函數(shù)的導(dǎo)數(shù)一、隱函數(shù)的導(dǎo)數(shù)二、由參數(shù)方程所確定的函數(shù)的導(dǎo)數(shù)三、相關(guān)變化率一、隱函數(shù)的導(dǎo)數(shù)?顯函數(shù)與隱函數(shù)形如y?f(x)的函數(shù)稱為顯函數(shù)?例如?y?sinx?y?lnx?ex都是顯函數(shù)?由方程F(x?y)?0所確的函數(shù)稱為隱函數(shù)?例如?方程x?y3?1?0確定的隱函數(shù)為y?31?x?把一個隱函數(shù)化成顯函數(shù)?叫做隱函數(shù)的顯化?一、隱函數(shù)的導(dǎo)數(shù)?隱函數(shù)的求導(dǎo)法把方程兩邊分別對x求導(dǎo)數(shù)?然后從所得的新的方程中把隱函數(shù)的導(dǎo)數(shù)解出?例1求由方程ey?xy?e?0所確定的隱函數(shù)y的導(dǎo)數(shù)?解方程中每一項對x求導(dǎo)得(ey)??(xy)??(e)??(0)?

2、?即ey?y??y?xy??0?yy從而y???(x?e?0)?x?ey提示:?(ey)??ey?y?????(xy)??y?xy????例2求由方程y5?2y?x?3x7?0所確定的隱函數(shù)y?f(x)在x?0處的導(dǎo)數(shù)y?

3、?x?0解法一把方程兩邊分別對x求導(dǎo)數(shù)得5y4?y??2y??1?21x6?0?1?21x6由此得y???5y4?2因為當(dāng)x?0時?從原方程得y?0?所以1?21x61y?

4、?

5、??x?04x?05y?22例2求由方程y5?2y?x?3x7?0所確定的隱函數(shù)y?f(x)在x?0處的導(dǎo)數(shù)y?

6、?x?0解法二把方程兩邊分別對x求導(dǎo)數(shù)得5y4?y

7、??2y??1?21x6?0?根據(jù)原方程?當(dāng)x?0時?y?0?將其代入上述方程得2y??1?0?從而y?

8、?0?5?x?0x2y23例例33?求橢圓??1在(2,3)處的切線方程?1692解把橢圓方程的兩邊分別對x求導(dǎo)?得x2?y?y??0??899x從而y????16y3當(dāng)x?2時?y?3?代入上式得所求切線的斜率23k?y?

9、???x?24所求的切線方程為33y?3??(x?2)?即3x?4y?83?0?241例例44．求由方程x?y?siny?0所確定的隱函數(shù)y2的二階導(dǎo)數(shù)?解方程兩邊對x求導(dǎo)?得dy1dy1??cosy??0?dx2dxdy2于是??d

10、x2?cosy上式兩邊再對x求導(dǎo)?得dyd2y?2siny??4siny?dx??dx2(2?cosy)2(2?cosy)3?對數(shù)求導(dǎo)法此方法是先在y?f(x)的兩邊取對數(shù)?然后用隱函數(shù)求導(dǎo)法求出y的導(dǎo)數(shù)?設(shè)y?f(x)?兩邊取對數(shù)?得lny?lnf(x)?兩邊對x求導(dǎo)?得1y??[lnf(x)]??yy??f(x)?[lnf(x)]??對數(shù)求導(dǎo)法適用于求冪指函數(shù)y?[u(x)]v(x)的導(dǎo)數(shù)及多因子之積和商的導(dǎo)數(shù)?例5求y?xsinx(x>0)的導(dǎo)數(shù)?解法一兩邊取對數(shù)?得lny?sinx?lnx?上式兩邊對x求導(dǎo)?得11y??cosx?lnx?sinx??y

11、xy??y(cosx?lnx?sinx?1?xsinx(cosx?lnx?sinx)?于是)xx解法二這種冪指函數(shù)的導(dǎo)數(shù)也可按下面的方法求???y?xsinx?esinx·lnx?y??esinx?lnx(sinx?lnx)??xsinx(cosx?lnx?sinx)?x(x?1)(x?2)例例66?求函數(shù)y?的導(dǎo)數(shù)?(x?3)(x?4)解先在兩邊取對數(shù)?得1lny?[ln(x?1)?ln(x?2)?ln(x?3)?ln(x?4)]?2上式兩邊對x求導(dǎo)?得111111y??(???)?y2x?1x?2x?3x?4y1111于是y??(???)?2x?1x?2x

12、?3x?4說明?嚴(yán)格來說?本題應(yīng)分x?4?x?1?2?x?3三種情況討論?但結(jié)果都是一樣的?二、由參數(shù)方程所確定的函數(shù)的導(dǎo)數(shù)?x??(t)設(shè)y與x的函數(shù)關(guān)系是由參數(shù)方程?確定的??y??(t)設(shè)x??(t)具有反函數(shù)t????(x)?且t????(x)與y??(t)構(gòu)成復(fù)合函數(shù)y??[???(x)]?若x??(t)和y??(t)都可導(dǎo)?則dydydtdy1??(t)??????dxdtdxdtdx??(t)dtdydy??(t)dydt即?或??dx??(t)dxdxdtdy??(t)若x??(t)和y??(t)都可導(dǎo)?則??dx??(t)?x?acost?例

13、例77?求橢圓?在相應(yīng)于t?點處的切線方程??y?bsint4dy(bsint)?bcostb解解?????cott?dx(acost)??asintadyb所求切線的斜率為????dxt?a4?2?2切點的坐標(biāo)為x0?acos?a?y0?bsin?b?42422b2切線方程為y?b??(x?a)?2a2即bx?ay?2ab?0?討論?已知x??(t),y??(t)?如何求y對x的二階導(dǎo)數(shù)y???提示?dy??(t)由x??(t)???dx??(t)d2yddyd??(t)dt?()?()dx2dxdxdt??(t)dx???(t)??(t)???(t)???

14、(t)1????2(t)??(t)??