cos+xy+对x求导

终诸亮2384有下列方程所确定的隐函数y在指定点的导数 -
熊蓝儿17654169399 ______ 1) 两边对x求导:y'e^y-y'sinx-ycosx=0 y'=ycosx/(e^y-sinx) 在点(0,1),y'=1/e2)两边对x求导:y'+e^y+xy'e^y=0 y'=-e^y/(1+xe^y) 在点(0,5),y'=-e^5

终诸亮23842、[cos(xy)]^3 +[sin(xy)]^3,求z关于x的偏导数(偏导数的符号不会打+ - +!) -
熊蓝儿17654169399 ______[答案] 直接把y看作是常数, 对x求导 z(x,y)=[cos(xy)]^3+[sin(xy)]^3 Zx(x,y)=3[cos(xy)]^2*[cos(xy)]'(xy)'+3[sin(xy)]^2*[sin(xy)]'(xy)' =3y[(sin(xy)^2(cos(xy))+cos(xy)^2(-sin(xy))]

终诸亮2384(6)题求导 -
熊蓝儿17654169399 ______ 两边对x求导,得cos(xy)·(y+xy')=1ycos(xy)+xcos(xy)y'=1xcos(xy)y'=1-ycos(xy)y'=[1-ycos(xy)]/xcos(xy)

终诸亮2384sinxy =xy 求dy -
熊蓝儿17654169399 ______[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...

终诸亮2384设函数y=f(x)由方程sin(xy)=x+y确定,求y'和dy. -
熊蓝儿17654169399 ______[答案] 对x求导,y是x的函数 所以cos(xy)*(xy)'=1+y' cos(xy)*(x'*y+x*y')=1+y' cos(xy)*(y+x*y')=1+y' ycos(xy)+xcos(xy)*y'=1+y' 所以y'=[ycos(xy)-1]/[1-xcos(xy)] 所以dy={[ycos(xy)-1]/[1-xcos(xy)]}dx

终诸亮2384隐函数求导问题,比较急隐函数求导时sin(xy)是变成什么,为什么这样变,我的理解是括号中这个数先把y看做系数对x求导再把x看做系数对y求导然后把它... -
熊蓝儿17654169399 ______[答案] sin(xy)' =cos(xy)*(xy)' =cos9xy)*(y+xy')

终诸亮2384求导sin(x+y)sin(x+y)这类型的该怎么求导? -
熊蓝儿17654169399 ______[答案] 如果对x求导,那么把y看成x的函数,整个函数是x的复合函数: (sin(x+y))'=cos(x+y)·(x+y)'=cos(x+y)·(x'+y')=(1+y')cos(x+y). 对y求导也类似.此例中x、y对称,上述结果中互换x、y即得到对y求导的结果:(1+x')cos(x+y),这里x'是x对y...

终诸亮2384计算下列隐函数的导数:cos(x+y)=xy,y' -
熊蓝儿17654169399 ______[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]

终诸亮2384sin x+cos xy+sin x^2+y^2=0这个隐函数能不能化成显函数? -
熊蓝儿17654169399 ______ sin x+cos xy+sin x^2+y^2=0这个隐函数是不能化成显函数的.如果要求导数y',只需将方程两端直接对x求导,即可.属于隐函数求导问题,不用化为显函数.

终诸亮2384微分方程题,怎样求得的y'cosx+ysinx=1? -
熊蓝儿17654169399 ______ ycosx+2∫(0,x)y(t)sintdt=x+1 两边对x求导 y'cosx-ysinx+2ysinx=1 y'cosx+ysinx=1