Answer
$y=-\frac{1}{2}x+7$
Work Step by Step
Step 1. Given the function $f(x)=\frac{8}{\sqrt {x-2}}$ and a point $(6,4)$, recall the formula for the differentiation of a function: $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$
Step 2. $f'(x)=\lim_{h\to0}\frac{\frac{8}{\sqrt {x+h-2}}-\frac{8}{\sqrt {x-2}}}{h}=8\lim_{h\to0}\frac{\sqrt {x-2}-\sqrt {x+h-2}}{h\sqrt {x+h-2}\sqrt {x-2}}=8\lim_{h\to0}\frac{(\sqrt {x-2}-\sqrt {x+h-2})(\sqrt {x-2}+\sqrt {x+h-2})}{h\sqrt {x+h-2}\sqrt {x-2}(\sqrt {x-2}+\sqrt {x+h-2})}=8\lim_{h\to0}\frac{x-2-x-h+2}{h\sqrt {x+h-2}\sqrt {x-2}(\sqrt {x-2}+\sqrt {x+h-2})}=\lim_{h\to0}\frac{-8}{\sqrt {x+h-2}\sqrt {x-2}(\sqrt {x-2}+\sqrt {x+h-2})}=\frac{-8}{(x-2)(2\sqrt {x-2})}=\frac{-4}{(x-2)^{3/2}}$
Step 4. At $x=6$, $f'(6)=\frac{-4}{(6-2)^{3/2}}=\frac{-4}{8}=-\frac{1}{2}$ which is the slope of the tangent line at point $(6,4)$
Step 5. The equation of the tangent line can be written as: $y-4=-\frac{1}{2}(x-6)$ or $y=-\frac{1}{2}x+7$
