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.

You can also consider taking the $\ln$ from both sides:
\begin{align*}
y = (x + \sqrt{1 + x^{2}})^{3/2} & \Rightarrow \ln(y) = \frac{3\ln(x + \sqrt{1 + x^{2}})}{2}\\\\
& \Rightarrow \frac{y'}{y} = \frac{3}{2(x + \sqrt{1 + x^{2}})}\times\left(1 + \frac{x}{\sqrt{1 + x^{2}}}\right)\\\\
& \Rightarrow y' = \frac{3y}{\sqrt{1 + x^{2}}}
\end{align*}
Hopefully this helps!
