<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] let $f(x) = x^a$
$f'(x) = a*x^{a-1}$
for example:
$f(x) = x^2$ -> $f'(x) = 2*x^1$
$f(x) = x^3$ -> $f'(x) = 3*x^2$
So...
if $f(x)=x^\frac{1}{2}$ what is $f'(x)$?
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