Taylor Expansion of $x\sqrt{x}$ at x=9 How can I go about solving the Taylor expansion of $x\sqrt{x}$ at x=9?
I solved the derivative down to the 5th derivative and then tried subbing in the 9 value for a using this equation 
$$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n.$$
Can someone walk me through the process (in neatly put-together LaTEX/Jax?)
I know I'm asking for a lot! Thanks. :)

Lets write down the first couple of derivatives first:
$f(x)=x\cdot \sqrt x$
$f'(x)=\frac{3\cdot\sqrt x }{2}$
$f''(x)=\frac{3}{4\cdot \sqrt x}$
You mentioned the taylor expansion in your opening post:
$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n.$
So your first three terms will be:
$\frac{f^{o}(9)}{0!}(x-9)^0=27$
$\frac{f^{1}(9)}{1!}(x-9)^1=\frac{9(x-9)}{2}$
$\frac{f^{2}(9)}{2!}(x-9)^2=\frac{1}{8}(x-9)^2$
(Just so you don't get confused $f^2=f''$)
Now it's your turn. Try to calculate the next couple of terms.
