第1个回答 2019-10-30
全微分法，
[(1/2)/(x^2+y^2)]*(2xdx+2ydy)=[(xdy-ydx)/x^2]/[1+(y/x)^2]
xdx+ydy=xdy-ydx，(y-x)dy=-(x+y)dx
y'=dy/dx=(x+y)/(x-y)本回答被提问者采纳
[(1/2)/(x^2+y^2)]*(2xdx+2ydy)=[(xdy-ydx)/x^2]/[1+(y/x)^2]
xdx+ydy=xdy-ydx，(y-x)dy=-(x+y)dx
y'=dy/dx=(x+y)/(x-y)本回答被提问者采纳
第2个回答 2019-10-30
ln√(x^2+y^2) = arctan(y/x)
(1/2)ln(x^2+y^2) = arctan(y/x)
两边求导
(1/2)[1/(x^2+y^2)] . (x^2+y^2)' ={ 1/[ 1+(y/x)^2] } .(y/x)'
(1/2)[1/(x^2+y^2)] . (2x+2y.y') ={ 1/[ 1+(y/x)^2] } .[ (xy'-y)/x^2]
(x+yy')/(x^2+y^2) = [x^2/(x^2+y^2) ] . [ (xy'-y)/x^2]
x+yy' = xy'-y
(x-y)y' = x+y
y' =(x+y)/(x-y)
(1/2)ln(x^2+y^2) = arctan(y/x)
两边求导
(1/2)[1/(x^2+y^2)] . (x^2+y^2)' ={ 1/[ 1+(y/x)^2] } .(y/x)'
(1/2)[1/(x^2+y^2)] . (2x+2y.y') ={ 1/[ 1+(y/x)^2] } .[ (xy'-y)/x^2]
(x+yy')/(x^2+y^2) = [x^2/(x^2+y^2) ] . [ (xy'-y)/x^2]
x+yy' = xy'-y
(x-y)y' = x+y
y' =(x+y)/(x-y)
第3个回答 2019-10-30
在作业帮助还有很多软件里面都能够搜索直接答案，所以说你在这问不如直接去网上搜索。