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求导数sin xy x y

见图

对x求导是ycos(xy).对y求导是xcos(xy)

求导计算过程如下: y=sin(x+y) 两边求导: y'=cos(x+y)*(1+y') y'=cos(x+y)+y'cos(x+y) y'[1-cos(x+y)]=cos(x+y) y'=cos(x+y)/[1-cos(x+y)].

∂Z/∂x= y*cos(xy) -2cos(xy)*sin(xy)*y = y*cos(xy) - y*sin(2xy) ∂Z/∂y= x*cos(xy) -2cos(xy)*sin(xy)*x = x*cos(xy) - x*sin(2xy)

解析如图所示

你好!可以如下图用隐函数求导法计算导数。经济数学团队帮你解答,请及时采纳。谢谢!

sin(xy)=x²y²+e^xy, 两边求导得到: cos(xy)(ydx+xdy)=2xy^2dx+2x^2ydy+e^(xy)(ydx+xdy) y[cos(xy)-e^(xy)]dx+x[cos(xy)-e^(xy)]dy=2xy^2dx+2x^2ydy x[cos(xy)-e^(xy)-2xy]dy=y[2xy-cos(xy)+e^(xy)]dx 所以: dy/dx=y[2xy-cos(xy)+e^(...

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)].

本题考查复合函数的求导法则 给题主重新算一遍 y = cos ( x+ y) y' = [ cos ( x + y )]' * ( x + y)' 链式法则,先求外面的函数的导数,再求里面函数的导数。 y' = -sin ( x + y ) * ( 1 + y') 函数求导法则,cos ( x+y)的导数是-sin(x+y),后面...

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