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. :)

When it's possible, it's wise to avoid explicit computation.
Note that
$$
x\sqrt{x} = [(x - 9)+9]\sqrt{x-9 + 9}=\\
9[(x/9 - 1)+1]\sqrt{9[x/9-1 + 1]}=\\
27[(x/9 - 1)+1]\sqrt{[x/9-1 + 1]}
$$
Let's define $z = x/9 - 1 = (x-9)/9$.  Find the Taylor expansion (in terms of $z$) for
$$
27[z + 1]\sqrt{z+1}
$$
By looking up the expansion for $\sqrt{z+1}$ in a table. Then, make the above substitution, and put the resulting series in the correct form.
