Question

求教一题导数题，第4题

Answer 1

Answer 2

arctan(y/x)=√(x^2+y^2)
两边求导
{ 1/[ 1+ (y/x)^2 ] } . [ (xy'-y) /x^2] = (x+ y.y')/√(x^2+y^2)
[1/(x^2+y^2)] (xy'-y) = (x+ y.y')/√(x^2+y^2)
xy'-y = (x+ y.y').√(x^2+y^2)
[x-y.√(x^2+y^2) ].y' = y + x.√(x^2+y^2)
y' = [y + x.√(x^2+y^2)]/[x-y.√(x^2+y^2) ]

Answer 3

arctan(y/x)=√(x^2+y^2)
y/x=tan√(x^2+y^2)

y=xtan√(x^2+y^2)
dy/dx=tan√(x^2+y^2)+xsec^2√(x^2+y^2)*(x+y*dy/dx)/√(x^2+y^2)
√(x^2+y^2)*dy/dx=√(x^2+y^2)*tan√(x^2+y^2)+xsec^2√(x^2+y^2)*(x+y*dy/dx)
[√(x^2+y^2)-xy*sec^2√(x^2+y^2)]dy/dx=√(x^2+y^2)*tan√(x^2+y^2)+x^2*sec^2√(x^2+y^2)
dy/dx=[√(x^2+y^2)*tan√(x^2+y^2)+x^2*sec^2√(x^2+y^2)]/[√(x^2+y^2)-xy*sec^2√(x^2+y^2)]