<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] 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
$$
{dy \over dx} = {3 y \over 2 \sqrt{1 + x^2}}
$$
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