Find equation of line normal to y= sqrt(x+2) over x

[cherica123]

Find equation of line normal to y= sqrt(x+2) over x

First thing i did was rewrite the sqrt so itd be (x+2)^.5 over x. I assume to find the derivative id use the division rule (low dee high minus high dee low, over the square of what's below!) Sorry if this will look messy but heres what i got:

(.5(x+2)^-.5)x - (x+2)^.5
over
x^2

even if my derivations were correct, i dont know where to go from here....if someone could check my work there and guide me down the right path to finish id really appreciate it....thanks!

final answer is y=(8/3)x-(13/3)

 

[pka]

\(\displaystyle y = \frac{{\sqrt {x + 2} }}{x}\quad \Rightarrow \quad y' = \frac{{ - \left( {x + 4} \right)}}{{2x^2 \sqrt {x + 2} }}\)

Therefore, any normal line has slope:

\(\displaystyle \frac{{2x^2 \sqrt {x + 2} }}{{\left( {x + 4} \right)}}.\)