#\frac { d } { d x } ( \frac { 4x ^ { 4} - 2x } { 4x ^ { 4} + 2x } ) =#
A common factor of #(2x)/(2x) = 1#:
#d/dx((2x^3-1)/(2x^3+1)) = #
Add zero to the numerator in the form if #+1 - 1#
#d/dx((2x^3+ 1 -1 -1)/(2x^3+1)) = #
#d/dx((2x^3+ 1 -2)/(2x^3+1)) = #
Break into two fractions:
#d/dx((2x^3+ 1)/(2x^3+1) -2/(2x^3+1)) = #
The first fraction becomes 1:
#d/dx(1 -2/(2x^3+1)) = #
Write the second fraction as a negative power:
#d/dx(1 -2(2x^3+1)^-1) = #
The derivative of 1 is 0:
#d/dx(-2(2x^3+1)^-1) = #
Use the chain rule:
#d/dx(f(g(x))) = (df)/(dg)(dg)/dx#
let #g = 2x^3+1#, then #f(g) = -2g^-1#, #(df)/(dg) = 2g^-2#, and #(dg)/dx = 6x^2#
#d/dx(-2(2x^3+1)^-1) = 2g^-2(6x^2)#
Reverse the substitution for g:
#d/dx(-2(2x^3+1)^-1) = 12x^2(2x^3+1)^-2#
#\frac { d } { d x } ( \frac { 4x ^ { 4} - 2x } { 4x ^ { 4} + 2x } ) =(12x^2)/(2x^3+1)^2#
