Answer:
Dy/Dx=-4x/(1+x²)²
Explanation:
The differential of 1_x² over 1+x²
First of all
1_x² over 1+x² = (1_x²) / (1+x²)
Let (1_x²) = u
Let (1+x²) = v
Differential = Dy/Dx
Dy/Dx of (1_x²) / (1+x²)
= (VDu/Dx -UDv/Dx)V²
u = (1-x²)
Du/Dx = -2x
(VDu/Dx) =(1+x²)(-2x)
V = 1+x²
Dv/Dx = 2x
UDv/Dx= (1-x²)(2x)
v² = (1+x²)²
Dy/Dx = ((1+x²)(-2x) - (1-x²)(2x))/(1+x²)²
Dy/Dx= ((-2x -2x³)-(2x-2x³))/(1+x²)²
Dy/Dx=( -2x -2x - 2x³ +2x³)/(1+x²)²
Dy/Dx=-4x/(1+x²)²