Let's perform the differentiation: ddx(xx2 + y2 -y2) ddx(xx2 + y2) + ddx(-y2) We differentiate u1 = x and v2 = x2 + y2 using: ddx(u1v2) = ddx(u1)·v2 -u1·ddx(v2)v22 Let's perform the differentiation u1: ddx(x) 1 Let's perform the differentiation v2: ddx(x2 + y2) ddx(x2) + ddx(y2) ℹddx(x2) = 2x‖ddx(y2) = 0 2x + 0 2x You can evaluate the resulted derivative (u1v2)': 1·(x2 + y2) -x·2x(x2 + y2)2 (x2 + y2) -x2·2(x2 + y2)2 -x2 + y2(x2 + y2)2 Then you can resume the derivation: ℹddx(xx2 + y2) = -x2 + y2(x2 + y2)2‖ddx(-y2) = 0 -x2 + y2(x2 + y2)2 + 0 -x2 + y2(x2 + y2)2 ▶Derivative: y2 -x2(x2 + y2)2