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.
 A: $$\frac{3}{2}\left(x+\sqrt{1+x²}\right)^\frac{1}{2}\left(1+\frac{x}{\sqrt{1+x²}}\right) \\  = \frac{3}{2}\left(x+\sqrt{1+x²}\right)^\frac{1}{2}\left(\frac{\sqrt{1+x²}+x}{\sqrt{1+x²}}\right) \\ = \frac{3}{2}\frac{\left(x+\sqrt{1+x²}\right)^\frac{3}{2}}{\sqrt{1+x²}} \\ = \frac{3}{2}\frac{y}{\sqrt{1+x²}} \\ = \frac{3y}{2(y-x)}$$  though you did not ask for the final step
A: 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
$$ \frac{dy}{dx} \ \  = \ \ \frac32 · [ \ x \ + \ \sqrt{1+x^2} \ ]^{1/2}   \ · \ \left( \ 1 \ + \ \frac{x}{\sqrt{1+x^2}} \ \right) \ \ = \ \ \frac32 · y^{1/3}   \ · \ \left( \frac{\sqrt{1+x^2} \ + \ x}{\sqrt{1+x^2}} \ \right) $$
$$ = \ \ \frac32 · y^{1/3}   \ · \   \frac{y^{2/3}  }{\sqrt{1+x^2}} \ \  = \ \   \frac{3 · y  }{2 · \sqrt{1+x^2}} \ \ . $$
A: A quick way of differentiation is to use logarithms on both sides of the equation:
$$
y = \left[ x + \sqrt{1 + x^2} \right]^{3 \over 2} \tag{1}
$$
We get
$$
\ln|y| = {3 \over 2} \ln\left| x + \sqrt{1 + x^2} \right| \tag{2}
$$
Differentiating both sides of (2) with respect to $x$, we get
$$
{1 \over y} \, {dy \over dx} = {3 \over 2} \ {1 + {2 x \over 2 \sqrt{1 + x^2}}
\over x + \sqrt{1 + x^2}}
$$
which can be simplified as
$$
{1 \over y} {dy \over dx} = {3 \over 2} \ {x + \sqrt{1 + x^2} \over
\sqrt{1 + x^2} \ [ x + \sqrt{1 + x^2} ]} \tag{3}
$$
Simplifying (3), we get
$$
{1 \over y}{dy \over dx} = {3 \over 2 \sqrt{1 + x^2}}
$$
Equivalently, we get
$$
{dy \over dx} = {3 y \over 2 \sqrt{1 + x^2}}
$$
A: 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!
