<s>[INST] Find the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $(9,3)$. Find the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $(9,3)$.
I did 
$$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{9+h}-3}{h} \frac{\sqrt(9+h)+3}{\sqrt(9+h)+3} = \frac{9+h-3}{h\sqrt{9+h}+3}= \frac{6}{\sqrt{9+h}+3}= \frac{6}{6}=1.$$
 [/INST] To follow up on Oria Gruber's answer:
If 
$y = \sqrt x, \tag{1}$
then
$x = y^2, \tag{2}$
so
$dx/dy = 2y = 2\sqrt x, \tag{3}$
whence
$dy/dx = (dx/dy)^{-1} = \frac{1}{2 \sqrt x}; \tag{4}$
thus when $x = 9$ we obtain
$dy/dx = 1/6, \tag{5}$
so the equation for the desired tangent line is
$y - 3 = \frac{1}{6}(x - 9). \tag{6}$
The derivative $dx/dy$ of $x$ with repect to $y$ may be easily had using the definition
$f'(x) = \lim_{h \to 0}((f(x + h) -f(x) / h), \tag{7}$
with which the OP seems familiar.
Hope this helps.  Cheerio,
and as always
Fiat Lux!!!
</s>