SquareRoot (Taylor Series)

The Taylor series of $f(x)$ centered at $a$ is given by $\displaystyle \sum_{n=0}^\infty \frac {f^{(n)}(a)}{n!}(x-a)^n$ , where $f^{(n)}$ is the $n$ th derivative of $f(x)$ . Therefore,

$\begin{aligned} \sqrt{x} \ \bigg|_{a=9} & = \sqrt{x} \ \bigg|_{x=9} + \frac d{dx} \sqrt{x} \ \bigg|_{x=9} (x-9) + \frac {d^2}{dx^2} \sqrt{x} \ \bigg|_{x=9} \frac {(x-9)^2}{2} + ... \\ & = \sqrt{x} \ \bigg|_{x=9} + \frac 1{2\sqrt{x}} \ \bigg|_{x=9} (x-9) - \frac 1{4 x\sqrt{x}} \ \bigg|_{x=9} \frac {(x-9)^2}{2} + ... \\ & = \boxed{3 + \dfrac 16 (x-9) - \dfrac 1{216}(x-9)^2 + ...} \end{aligned}$