cos括弧xy怎么求导

凤将沸1624求导cos=x的显函数的导数只求一阶导数, -
苗爽垄13188404674 ______[答案] 对x求导 -sinxy*(xy)'=1 (xy)'=x'*y+x*y'=y+x*y' 所以y+x*y'=-1/sin(xy) y'=-[1/sin(xy)+y]/x

凤将沸1624y=cos(xy)+x.请问怎么求导不是求导,是微分,不好意思~ -
苗爽垄13188404674 ______[答案] 把y看作因变量,有 dy=d[cos(xy)]+dx dy=-sin(xy)(ydx+xdy)+dx 化简得dy/dx=[1-ysin(xy)]/[1+xsin(xy)]

凤将沸1624由cos(xy)=x+y确定y是关于x的函数,求y'.能不能把刚才的题过程也给我写下,不懂啊 -
苗爽垄13188404674 ______[答案] cos(xy)=x+y 两边分别对x求导: -sin(xy)*(y+xy′)=1+y′ y′=-[1+ysin(xy)]/[xsin(xy)+1] ========= 左边对cos求导:-sin(xy) 再对xy求导:y+xy′ 右边对x求导:1+y′ 左边两项相乘,最后解出y′

凤将沸1624(e^y)cos x = 6 + sin(xy) 这个求导怎么求啊...要详细过程RT在线等 -
苗爽垄13188404674 ______[答案] 对x求导是吧, (e^y)cos x = 6 + sin(xy), 所以对等式两边求导得到, (e^y)' *cosx +(e^y) *(cosx)' = [sin(xy)]' 显然(e^y)'= (e^y) *y',(cosx)'= -sinx,[sin(xy)]'= cos(xy) *(xy)'=cos(xy) *(y+xy') 于是 (e^y) *y' *cosx - (e^y) *sinx= cos(xy) *(y+xy'), 那么化简得...

凤将沸1624求方程所确啶的隐函数的导数:e^x - e^y=sin(xy),求y'(0). -
苗爽垄13188404674 ______[答案] 隐函数求导的一般步骤: 1.等式两边同时对X求导,把Y看成X的复合函数 2.Y`(X)放到一边,其余的放到另一边,可求出 这个题目: 1.两边求导数: (e^x)`-(e^y)=[sin(xy)]` e^x-(e^y)*y`=cos(xy)*(xy)` e^x-(e^y)*y`=cos(xy)*(y+xy`) e^x-ycos(xy)*y=...

凤将沸1624sinxy =xy 求dy -
苗爽垄13188404674 ______[答案] sin(xy) =xy 典型的隐函数求导,两边对x求导即可:cos(xy)*(y+x*dy/dx)=y+x*dy/dx展开cos(xy)*y+cos(xy)*x*dy/dx=y+x*dy/dx两边乘以dxcos(xy)*y*dx+cos(xy)*x*dy=y*dx+x*dy移项cos(xy)*y*dx-y*dx=x*dy-cos(xy)*x*dydy...

凤将沸1624隐函数为什么可这样求导 求导依据 -
苗爽垄13188404674 ______ 隐函数求导的依据是,假定该函数可导,把隐函数的式子左、右边均看成一个整体的函数,并且把函数中的y看做是还有下一级函数的复合函数y(x),然后利用复合函数的求导法则进行求导,最后把y'(x)解出来,用含x、y的式子表达.例如:sin(xy)=2x+y^2 求隐函数y的导数.假定y可导,此时,把函数的左右均看做是整体,而y是下一级的复合函数,于是,利用复合函数的求导法则求导,得 cos(xy)*(y+xy')=2+2y*y' 解上式,得 y'=[2-ycos(xy)]/[xcos(xy)-2y]

凤将沸1624求下列隐函数的一阶导数 y' 1、cos(xy)=x+y 2、y=tan(x+y) 我算的答案总是跟标准的不一样,只好求助了 -
苗爽垄13188404674 ______ 1、cos(xy)=x+y [cos(xy)]′=(x+y)′-sin(xy)*(xy)′=1+y′-sin(xy)*(x′y+xy′)=1+y′-sin(xy)*(y+xy′)-1-y′=0 [xsin(xy)+1]y′=-ysin(xy)-1 y′=-[ysin(xy)+1]/[xsin(xy)+1]2、y=tan(x+y) y′=sec²(x+y)*(x+y)′=sec²(x+y)*(1+y′) [1-sec²(x+y)]y′=sec²(x+y) y′=sec²(x+y)/[1-sec²(x+y)]=1/[cos²(x+y)-1]=-1/sin²(x+y)=-csc²(x+y)

凤将沸1624计算下列隐函数的导数:cos(x+y)=xy,y' -
苗爽垄13188404674 ______[答案] cos(x+y)=xy两边求导得: -(1+y')*sin(x+y)=y+xy' -sin(x+y)-y=y'*[sin(x+y)+x] y'=[-sin(x+y)-y]/[sin(x+y)+x]

凤将沸1624一道简单的数学偏导数
苗爽垄13188404674 ______ f对x的偏导 2cos(xy)*(-sinxy)*y= - ysin(2xy)