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?

• Check whether the above edited function is correct, @Hello. Aug 27, 2014 at 23:44
• That's a lot of tedious algebra work for one multiple choice question... Aug 27, 2014 at 23:45
• @Timbuc yes thank you. Aug 27, 2014 at 23:45
• @Nameless yeah i know haha Aug 27, 2014 at 23:45
• Ok @Hello, now you edit your post and make it correct copying what I edited. Aug 27, 2014 at 23:47

$$f(x)=2x \sqrt{x^2+8}+2 \Rightarrow f'(x)=2\sqrt{x^2+8}+\frac{x}{\sqrt{x^2+8}}2x=2\sqrt{x^2+8}+\frac{2x^2}{\sqrt{x^2+8}}=\frac{2(x^2+8)+2x^2}{\sqrt{x^2}8}=\frac{4x^2+16}{\sqrt{x^2+8}}$$

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:

$$y-y_1=m(x-x_1) \Rightarrow y-2=-\frac{\sqrt{2}}{8}(x-0) \Rightarrow 8y-16=-\sqrt{2}x \Rightarrow 8y+\sqrt{2}x=16$$

$$8y+\sqrt{2}x=16 \Rightarrow 8 \sqrt{2}y+2x=16 \sqrt{2} \Rightarrow 4 \sqrt{2}y+x=8 \sqrt{2}$$

• Shouldn't the slope be negative when you said "The slope of the normal line is..." you wrote -1/f'(0) but never actually made it negative Aug 27, 2014 at 23:57
• Oh, sorry! I corrected it! Aug 28, 2014 at 0:00
• I understand how you got that and I got it too. But it's a multiple choice question (check the OP where I posted the link (#9)). Any ideas how to turn that into that? Aug 28, 2014 at 0:01
• I edited my answer! Therefore, it is $b$. Aug 28, 2014 at 0:06

$$f'(x)=\left(2x\sqrt{x^2+8}+2\right)'=2\sqrt{x^2+8}+\frac{2x^2}{\sqrt{x^2+8}}=\frac{4x^2+16}{\sqrt{x^2+8}}$$

So

$$f'(0)=\frac{16}{\sqrt8}=\frac{8}{\sqrt2}=4\sqrt2\implies-\frac1{f'(0)}=-\frac1{4\sqrt{2}}$$

Try now to take it from here.

• Wait before I continue, I had the right numbers? lol Aug 27, 2014 at 23:47
• I can't be sure until you first edit your question and make it clear... Aug 27, 2014 at 23:48
• Yeah I actually got that far already. I got $-1/(2\sqrt{8})$ which is the same thing Aug 27, 2014 at 23:51