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偏导数与全微分习题1.设yxyxyxfarcsin)1(),(,求)1,(xfx。2.习题817题。3.设0001sin),(222222yxyxyxyyxf,考察f(x,y)在点(0,0)的偏导数。4.考察0001sin),(222222yxyxyxxyyxf在点(0,0)处的可微性。5.证明函数0001sin)(),(22222222yxyxyxyxyxf在点(0,0)连续且偏导数存在,但偏导数在(0,0)不连续,而f(x,y)在点(0,0)可微。1.设yxyxyxfarcsin)1(),(,求)1,(xfx。yyxyxyyxfx1)(2111)1(1),(21∴1)1,(xfx。2.习题817题。17.设22)()(lnbyaxz(a,b为常数),证明02222yzxz。先化简函数))()ln((2122byaxz,2222)()()()()()(221byaxaxbyaxaxxz,2222)()()()()()(221byaxbybyaxbyyz,22222222))()(()(2)()(byaxaxbyaxxz22222))()(()()(byaxaxby,22222222))()(()(2)()(byaxbybyaxyz22222))()(()()(byaxbyax,∴02222yzxz。3.设0001sin),(222222yxyxyxyyxf,考察f(x,y)在点(0,0)的偏导数。由偏导数定义可知00lim)0,0()0,(lim)0,0(00xxxxfxff,2001sinlim)0,0(),0(lim)0,0(yyfyffyyy不存在。4.考察0001sin),(222222yxyxyxxyyxf在点(0,0)处的可微性。由偏导数定义可知0)0,0()0,(lim)0,0(0xfxffxx,0)0,0(),0(lim)0,0(0yfyffyy,则dz=0,22)()(1sin)0,0(),(yxyxfyxfdzf要讨论在(0,0)点可微性,即讨论极限dzf0lim是否趋于0,0)()()()(1sinlimlim222200yxyxyxdzf,这是因为22222222)()()()(21|)()()()(1sin|yxyxyxyxyx22)()(21yx∴f(x,y)在点(0,0)处的可微4.证明函数0001sin)(),(22222222yxyxyxyxyxf在点(0,0)连续且偏导数存在,但偏导数在(0,0)不连续,而f(x,y)在点(0,0)可微。(1)连续|1sin)(||)0,0(),(|2222yxyxfyxf||22yx,故f(x,y)在(0,0)点连续;(2)偏导数存在由偏导数定义0||1sin)(lim)0,0()0,(lim)0,0(200xxxxfxffxxx同理0)0,0(xf,偏导数存在;(3)偏导数在(0,0)点不连续当022yx时2222221cos1sin2),(yxyxxyxxyxfx,而220021cos||221sin2lim),(limxxxxxyxfxyxyxxx极限不存在,故),(yxfx在(0,0)处不连续;同理,),(yxfy在(0,0)处不连续;(4)可微由(2)可知:dz=0,)0,0(),(fyxfdzf2222)()(1sin))()((yxyx,22222200)()()()(1sin))()((limlimyxyxyxdzf0)()(1sin])()[(lim2221220yxyx,∴f(x,y)在(0,0)点可微。
本文标题:偏导数与全微分习题
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