<s>[INST] Find the equation of normal line to the graph of given function 
Give the equation of the normal line to the graph of
  $$y =  2x \sqrt{x^2+8} + 2$$
  at the point $(0,2)$

What I've done so far is:
Taken the derivative and got
$$(2x^2)/\sqrt{ x^2+8} +  2\sqrt{x^2+8}$$
I have no idea if this is right, it was pretty hard to get the derivative of that. Before I go on any further, is this derivative right?
Source: http://online.math.uh.edu/apcalculus/exams/AP_AB_version1_1.htm - #9
 [/INST] $$f(x)=2x \sqrt{x^2+8}+2 \Rightarrow f'(x)=2\sqrt{x^2+8}+\frac{x}{\sqrt{x^2+8}}2x=2\sqrt{x^2+8}+\frac{2x^2}{\sqrt{x^2+8}}=\frac{2(x^2+8)+2x^2}{\sqrt{x^2}8}=\frac{4x^2+16}{\sqrt{x^2+8}}$$
At the point $(0,2)$, $f'(0)=\frac{16}{\sqrt{8}}=\frac{16}{2 \sqrt{2}}=\frac{8}{\sqrt{2}}$.
The slope of the normal line is $\frac{-1}{f'(0)}=-\frac{\sqrt{2}}{8}$.
Therefore, the equation of the normal line at the point $(0,2)$ is the following:
$$y-y_1=m(x-x_1) \Rightarrow y-2=-\frac{\sqrt{2}}{8}(x-0) \Rightarrow 8y-16=-\sqrt{2}x \Rightarrow 8y+\sqrt{2}x=16$$
$$8y+\sqrt{2}x=16 \Rightarrow 8 \sqrt{2}y+2x=16 \sqrt{2} \Rightarrow 4 \sqrt{2}y+x=8 \sqrt{2}$$
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