ln[根号(x^2+y^2)] =arctany/x 求dy

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ln[根号(x^2+y^2)]=arctany/x求dyln[根号(x^2+y^2)]=arctany/x求dyln[根号(x^2+y^2)]=arctany/x求dy1/2*ln(x^2+y^2)=

ln[根号(x^2+y^2)] =arctany/x 求dy
ln[根号(x^2+y^2)] =arctany/x 求dy

ln[根号(x^2+y^2)] =arctany/x 求dy
1/2*ln(x^2+y^2)=arctany/x两边对x求导,得
1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(y/x)^2]*(y'*x-y)/x^2
化简得
y'=(x+y)/(x-y)
则dy=(x+y)/(x-y)*dx

‘=-1 当sinx<0时 y’=1 2. arctany/x=ln√(x +y ) (这里题目真变态! y=arcsin(cosx) y'=arcsin(cosx)' =1/√(1-cos^2x)