Claire wants to find the derivative of $h(x)=18\sqrt{x}$ at the point $x=9$.
Her table below shows the slopes of the secant lines to the graph of $h$ between the points $(9,h(9))$ and $(x,h(x))$ for $x$-values that get increasingly closer to $9$:
$x$|Slope of secant line, $\dfrac{h(x)-h(9)}{x-9}$
:- | :- | :-
$8.9$| $3.0084$
$8.99$| $3.0008$
$8.999$| $3.0001$
$9.001$| $2.9999$
$9.01$| $2.9992$
$9.1$| $2.9917$
**From the table, what does the derivative of $h(x)=18\sqrt{x}$ at $x=9$ appear to be?**
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