Simplifying a derivative of a function involving roots. I am working on a question where I need to differentiate $$y=\{x+\sqrt{1+x²}\}^\frac{3}{2}\tag1$$
Using the chain rule I have found the first derivative $$\frac{dy}{dx} =\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{1}{2}\{1+\frac{x}{\sqrt{1+x²}}\}\tag2$$
Which can be written as
$$\frac{dy}{dx} = \frac{3y}{2\{x+\sqrt{1+x²}\}}\{1+\frac{x}{\sqrt{1+x²}}\}\tag3$$
This is as far as I have got in terms of simplifying. The solution says that $(2)$ is equal to $(4)$ $$\frac{\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{3}{2}}{\sqrt{1+x²}}=\frac{3y}{2\sqrt{1+x^2}}\tag4$$
How do you get from $(2)$ to $(4)$? I have included $(3)$ in order to show my workings.

Since it is desired to have an expression for the derivative that includes $ \ y \ \ , $ we will want to keep in mind that
$$ y \ \ = \ \ [ \ x \ + \ \sqrt{1+x^2} \ ]^{3/2} \ \ \Rightarrow \ \ x \ + \ \sqrt{1+x^2} \ \ = \ \ y^{2/3} $$
and thus $ \ [ \ x \ + \ \sqrt{1+x^2} \ ]^{1/2} \   =   \ y^{1/3} \ \ . $
The first derivative can then be written as
$$ \frac{dy}{dx} \ \  = \ \ \frac32 · [ \ x \ + \ \sqrt{1+x^2} \ ]^{1/2}   \ · \ \left( \ 1 \ + \ \frac{x}{\sqrt{1+x^2}} \ \right) \ \ = \ \ \frac32 · y^{1/3}   \ · \ \left( \frac{\sqrt{1+x^2} \ + \ x}{\sqrt{1+x^2}} \ \right) $$
$$ = \ \ \frac32 · y^{1/3}   \ · \   \frac{y^{2/3}  }{\sqrt{1+x^2}} \ \  = \ \   \frac{3 · y  }{2 · \sqrt{1+x^2}} \ \ . $$
