Answer
$f'(x) = \frac{-4}{(x-2)^{\frac{3}{2}}}$
Tangent line:
$y = 4 -\frac{1}{2}(x-6)=-\frac{1}{2}x+7$
Work Step by Step
$f(x) =\frac{8}{\sqrt (x-2)}$
Using the derivative property:
$f'(x) = \lim\limits_{h \to 0} \frac{\frac{8}{\sqrt ((x+h)-2)} -\frac{8}{\sqrt (x-2)} }{h}$
$f'(x) = \lim\limits_{h \to 0}\frac{-8}{(\sqrt ((x+h)-2))(\sqrt (x-2))({\sqrt ((x+h)-2)} +\sqrt (x-2))}$
$f'(x) = \frac{-4}{(x-2)^{\frac{3}{2}}}$
Slope at x = 6:
$slope = f'(6) = -\frac{1}{2}$
So, the equation of tangent line is:
$\frac{y-4}{x-6} = -\frac{1}{2}$
$y = 4 -\frac{1}{2}(x-6)=-\frac{1}{2}x+7$
