<s>[INST] 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.
 [/INST] $$\frac{3}{2}\left(x+\sqrt{1+x²}\right)^\frac{1}{2}\left(1+\frac{x}{\sqrt{1+x²}}\right) \\  = \frac{3}{2}\left(x+\sqrt{1+x²}\right)^\frac{1}{2}\left(\frac{\sqrt{1+x²}+x}{\sqrt{1+x²}}\right) \\ = \frac{3}{2}\frac{\left(x+\sqrt{1+x²}\right)^\frac{3}{2}}{\sqrt{1+x²}} \\ = \frac{3}{2}\frac{y}{\sqrt{1+x²}} \\ = \frac{3y}{2(y-x)}$$  though you did not ask for the final step
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