cosxy对y偏导

阙南齐3126y/(x+y)²对x求偏导数,最好有过程 -
东话彦13263285362 ______ 对x 求偏导数的时候,就将y视为常数,那么y/(x+y)² 即 y *(x+y)^(-2) 对x 求偏导数得到-2y *(x+y)^-3

阙南齐3126cosxy=x的导数和y=cos(x+y)的导数怎么求,详细过程,谢谢! -
东话彦13263285362 ______ cosxy=x -sinxy*(y+xy')=1 y+xy'=-cscxy y'=-(cscxy+y)/x. y=cos(x+y) y'=-sin(x+y)*(1+y') y'[1+sin(x+y)]=-sin(x+y) y'=-sin(x+y)/[1+sin(x+y)].

阙南齐3126.一平面流场速度为ux = k sin(xy),uy = ─ sin(xy).若保证流场连续,则k值为 -
东话彦13263285362 ______ 1. ux = k sin(xy ) 对x求偏导 为 kycos(xy)2. uy = ─ sin(xy) 对y求偏导 为 -xcos(xy)3. 保证流场连续 则 kycos(xy) -xcos(xy)=0 k=x/y

阙南齐3126设f(x,y)=cscxy,则f'x(π/2,1)= -
东话彦13263285362 ______ cscxy对x求偏导数 得到y *(-cosxy)/sin²xy 代入x=π/2,y=1 于是f'x(π/2,1)=(-cosπ/2)/sin²π/2 =0/1=0,选择B

阙南齐3126求导:已知y=cos(xy),求y的一阶导数(用隐函数求导) -
东话彦13263285362 ______ 解:对等式两边求导,得 y'=-sin(xy)*(y+xy') y'=-ysin(xy)/[xsin(xy)+1]

阙南齐3126设u=(e^xy)(cos(x+y^2)),求du -
东话彦13263285362 ______ 对x偏导du/dx=ye^xy(cos(x+y^2))+e^xy(-sin(x+y^2)) 偏导的那个a打不出来 对y偏导du/dy=xe^xy(cos(x+y^2))+e^xy(-sin(x+y^2))2y du={ye^xy(cos(x+y^2))+e^xy(-sin(x+y^2))}dx+{xe^xy(cos(x+y^2))+e^xy(-sin(x+y^2))2y}dy 里面的小项没有整理...

阙南齐3126求偏导数(高数下册)
东话彦13263285362 ______ 3) 对x求导 =f1(e^xy,x²-y²)·e^xy·y+f2(e^xy,x²-y²)·2x 对y求导 =f1(e^xy,x²-y²)·e^xy·x+f2(e^xy,x²-y²)·(-2y) 4) 对x求导 =f1(sinxy,x-y²)·cosxy·y+f2(sinxy,x-y²) 对y求导 =f1(sinxy,x-y²)·cosxy·x+f2(sinxy,x-y²)·(-2y)

阙南齐3126sinxy=1,那么相等于多少? -
东话彦13263285362 ______ xy=2kπ+π/2 ,k=Z

阙南齐3126sinxy+e^x=y^2 求dy/dx 解题中 Fx=ycosxy+e^x 中的y哪里来的?不是F对x的偏导应该是cosxy+e^x? -
东话彦13263285362 ______[答案] 解; sinxy对x求导是(sinxy)'(xy)'=(cosxy)(y)=ycosxy ∴是Fx=ycosxy+e^x

阙南齐3126Z= e^xy+yx^2,求dz -
东话彦13263285362 ______[答案] dz=z对x偏导*dx+z对y偏导*dy=[y*(e^xy)+2xy]*dx+[x*(e^xy)+x^2]*dy