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

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Given y= ln(1+2x+2x^2)

We need to find y'.

We will use the chain rule to solve.

Let y= ln u such that:

u= 1+2x+2x^2 ===> u' = 2+4x

Now we will substitute:

==> y= ln u

==> y' = (ln u)' = 1/u * u'

Now we will substitute:

==> y' = 1/(1+2x+2x^2) * (2+ 4x)

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

**==> y' = (2+4x)/(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)**