Answer
$y=\frac{1}{6}x+\frac{3}{2}$
Work Step by Step
Step 1. Given the function $f(x)=\sqrt x$ and $x=9$, we have $f(9)=\sqrt 9=3$, so the point is $(9,3)$
Step 2. The slope of the tangent line is the derivative of the function at this point:
$f'(9)=\lim_{h\to0}\frac{f(9+h)-f(9)}{h}=\lim_{h\to0}\frac{\sqrt {9+h}-3}{h}=\lim_{h\to0}\frac{(\sqrt {9+h}-3)(\sqrt {9+h}+3)}{h(\sqrt {9+h}+3)}=\lim_{h\to0}\frac{9+h-9}{h(\sqrt {9+h}+3)}=\lim_{h\to0}\frac{1}{\sqrt {9+h}+3}=\frac{1}{6}$
Step 3. Knowing the slope $\frac{1}{6}$ and the passing point $(9,3)$, we can write the equation of the tangent line as: $y-3=\frac{1}{6}(x-9)$ or $y=\frac{1}{6}x+\frac{3}{2}$
