Question

求全微分！谢谢

Answer 1

f(x,y) = arctan[(x+y)/(1+xy)]
f'(x,y) = {1/[1+(x+y)^2/(1+xy)^2]} [(1+xy-(x+y)y]/(1+xy)^2
= (1-y^2)/[(x+y)^2+(1+xy)^2],
同理 f'(x,y) = (1-x^2)/[(x+y)^2+(1+xy)^2],
当 |x|， |y| 很小时 f'(x,y) ≈ 1， f'(x,y) ≈ 1 ，
⊿f = x+y, f(x,y) ≈ f(0,0) + x+y = x+y.

Answer 2

dz=d(x^2y^2)/(x^2+y^2)^(3/2)+x^2y^2*d[1/(x^2+y^2)^(3/2)]
=(2xy^2dx+2x^2ydy)/(x^2+y^2)^(3/2)
+x^2y^2*(-3/2)(2xdx+2ydy)/(x^2+y^2)^(5/2)
=[(2xy^2dx+2x^2ydy)(x^2+y^2)-x^2y^2(3xdx+3ydy)]/(x^2+y^2)^(5/2)
=[(2x^3y^2+2xy^4-3x^3y^2)dx+(2x^4y+2x^2y^3-3x^2y^3)dy]/(x^2+y^2)^(5/2)
=[(2xy^4-x^3y^2)dx+(2x^4y-x^2y^3)dy]/(x^2+y^2)^(5/2),x^2+y^2≠0.