Differentiate y= ln(1+2x+2x^2)?

We'll apply the chain rule to determine the result of the
first derivative:

dy/dx =
d[ln(1+2x+2x^2)]/dx

dy/dx =
[d(1+2x+2x^2)/dx]/(1+2x+2x^2)

dy/dx = [(d/dx)(1) +
(d/dx)(2x) + (d/dx)(2x^2)]/(1+2x+2x^2)

dy/dx = (0 + 2 +
4x)/(1+2x+2x^2)

dy/dx = (4x + 2)/(2x^2 + 2x +
1)

dy/dx = 2(2x + 1)/(2x^2 + 2x +
1)

The result of differentiating y
is:

dy/dx = 2(2x + 1)/(2x^2 +
2x + 1)