## University Calculus: Early Transcendentals (3rd Edition)

$f'(x) = \frac{-4}{(x-2)^{\frac{3}{2}}}$ Tangent line: $y = 4 -\frac{1}{2}(x-6)=-\frac{1}{2}x+7$
$f(x) =\frac{8}{\sqrt (x-2)}$ 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: $\frac{y-4}{x-6} = -\frac{1}{2}$ $y = 4 -\frac{1}{2}(x-6)=-\frac{1}{2}x+7$