The problem asks to find the derivative $y'$ of the function $y = \frac{2x}{\sqrt{x^2 + 2x + 2}}$. We are given the form of the answer $y' = \frac{[1](x + [2])}{(x^2 + 2x + 2)^{\frac{3}{2}}}$ and we need to find the values for [1] and [2].

1. Problem Description

The problem asks to find the derivative $y'$ of the function $y = \frac{2x}{\sqrt{x^2 + 2x + 2}}$. We are given the form of the answer $y' = \frac{[1](x + [2])}{(x^2 + 2x + 2)^{\frac{3}{2}}}$ and we need to find the values for [1] and [2].

2. Solution Steps

We will use the quotient rule to find the derivative of $y$. The quotient rule states that if $y = \frac{u}{v}$, then $y' = \frac{u'v - uv'}{v^2}$.

Here, $u = 2x$ and $v = \sqrt{x^2 + 2x + 2} = (x^2 + 2x + 2)^{\frac{1}{2}}$.

We have $u' = 2$.

To find $v'$, we use the chain rule: $v' = \frac{1}{2}(x^2 + 2x + 2)^{-\frac{1}{2}}(2x + 2) = \frac{x+1}{\sqrt{x^2 + 2x + 2}}$.

Now, applying the quotient rule:

$y' = \frac{2\sqrt{x^2 + 2x + 2} - 2x \cdot \frac{x+1}{\sqrt{x^2 + 2x + 2}}}{(\sqrt{x^2 + 2x + 2})^2}$

$y' = \frac{2(x^2 + 2x + 2) - 2x(x+1)}{(x^2 + 2x + 2)\sqrt{x^2 + 2x + 2}}$

$y' = \frac{2x^2 + 4x + 4 - 2x^2 - 2x}{(x^2 + 2x + 2)^{\frac{3}{2}}}$

$y' = \frac{2x + 4}{(x^2 + 2x + 2)^{\frac{3}{2}}}$

$y' = \frac{2(x + 2)}{(x^2 + 2x + 2)^{\frac{3}{2}}}$

Comparing this with the given form $y' = \frac{[1](x + [2])}{(x^2 + 2x + 2)^{\frac{3}{2}}}$, we can see that $[1] = 2$ and $[2] = 2$.

3. Final Answer

The value of [1] is 2, and the value of [2] is

2.