求z=sin(xy)+cos（的平方）(xy) 函数的偏导数~~~求具体步骤 求下列函数的偏导数：z=sin(xy)+cos²(...
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dz/dx=ycos(xy)-2ycos(xy)sin(xy)=ycos(xy)[1-2sin(xy)]
dz/dy=xcos(xy)-2xcos(xy)sin(sy)=xcos(xy)[1-2sin(xy)]
令 u = x y, əu/əx = y, əu/əy = x
z = sinu + cos²u
əz/əx = [cosu - sin(2u)] * əu/əx = y [cos(x y) - sin(2x y)]
əz/əy = [cosu - sin(2u)] * əu/əy = x [cos(x y) - sin(2x y)]
绛旓細1.鍏堟眰涓闃跺亸瀵硷紝姣斿瀵箈姹備竴闃跺亸瀵硷紝瀵箈姹傚亸瀵兼暟灏辨妸y鍜寊瑙嗕负甯告暟锛岀粨鏋滀负sin(xy)'(xy)'cos(yz)+sin(xy)cos(yz)'(yz)'=cos(xy)*y*cos(yz)-sin(xy)*sin(yz)*0=cos(xy)*y*cos(yz) //娉ㄦ剰(xy)' 瀵箈姹傚亸瀵兼椂缁撴灉涓簓,鍥犱负姝ゆ椂y涓哄父鏁般傜劧鍚庡啀瀵箉,z姹鍋忓锛屽師鐞嗘槸涓...
绛旓細z=sin(xy)az/ax=cos(xy) *y=ycos(xy)az/ay=cos(xy) *x=xcos(xy)
绛旓細鎴戠殑 姹傚亸瀵兼暟:(1)z=sin(xy)+cos(x+y)(2)z=(1+xy)鐨剏娆℃柟 鎴戞潵绛 1涓洖绛 #鐑# 璇ヤ笉璇ヨ瀛╁瓙寰堟棭瀛︿範浜烘儏涓栨晠?鐧惧害缃戝弸af34c30f5 2017-06-06 路 TA鑾峰緱瓒呰繃4.3涓囦釜璧 鐭ラ亾澶ф湁鍙负绛斾富 鍥炵瓟閲:1.8涓 閲囩撼鐜:65% 甯姪鐨勪汉:5087涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 ...
绛旓細z瀵筙鐨勫亸瀵兼暟=Ycosxy+2cos(xy)(-sinxy*(-y)) 瀵筙姹傚亸瀵兼暟鏃讹紝鎶奩鐪嬫垚甯告暟锛屽啀瀵筙杩涜姹傚灏辫锛屽Y姹傚亸瀵兼暟涔熶竴鏍
绛旓細鍘熷嚱鏁 z=sin(x*y)+(cos(x*y))^2 əz/əx =y*cos(x*y) - 2*y*cos(x*y)*sin(x*y)əz/əy =x*cos(x*y) - 2*x*cos(x*y)*sin(x*y)
绛旓細dz=cos(xy)ydx+cos(xy)xdy=(ydx+xdy)cos(xy)锛岄塂
绛旓細鐢眤/x=ycos(xy),z/y=xcos(xy)鈫抎z=ycos(xy)dx + xcos(xy)dy.
绛旓細dz=cos(xy)xdy+cos(xy)ydx
绛旓細瑙ｉ杩囩▼濡備笅锛歞z/dx=ycosxy dz/dy=xcosxy d^2z/dx^2=y^2cosxy d^2z/dy^2=x^2cosxy =cosxy-xcosy 鎬ц川锛氫簩闃跺鏁版槸涓闃跺鏁扮殑瀵兼暟銆備粠鍘熺悊涓婄湅锛屽畠琛ㄧず涓闃跺鏁扮殑鍙樺寲鐜囷紱浠庡浘褰笂鐪嬶紝瀹冨弽鏄犵殑鏄嚱鏁板浘鍍忕殑鍑瑰嚫鎬с備簩闃跺鏁板彲浠ュ弽鏄犲浘璞＄殑鍑瑰嚫銆備簩闃跺鏁板ぇ浜0锛屽浘璞′负鍑癸紱浜岄樁瀵兼暟...
绛旓細瑙ｏ細鐢眤/x=ycos(xy)锛寊/y=xcos(xy)鈫抎z=ycos(xy)dx + xcos(xy)dy銆傞夯鐑﹂噰绾筹紝璋㈣阿!
