<s>[INST] Maclurin series for $e^{\sqrt{x^2+1}}$ i am giving $e^{\sqrt{x^2+1}}$ and asked to find the Maclaurin series for this term.

Here is my solution:
let $u=\sqrt{x^2+1}$, and given that we know that Maclaurin series for $e^x= 1+x+\frac{x^2}{2!} ...$ 
then: $$e^u= 1+u+\frac{u^2}{2}+...$$
hence: 
$$e^{\sqrt{x^2+1}}=1+\sqrt{x^2+1}+\frac{x^2+1}{2}+...$$
Am I doing it right? please help
 [/INST] You can get the first terms of the series computing the derivatives of $f(x)=e^{\sqrt{x+1}}$. You get:$$\begin{align}f(x)&=e^{\sqrt{x+1}}\\f'(x)&=\frac{e^{\sqrt{x+1}}}{2\sqrt{x+1}}\\f''(x)&=\frac{e^{\sqrt{x+1}} \left(\sqrt{x+1}-1\right)}{4(x+1)^{3/2}}\\f^{(3)}(x)&=\frac{e^{\sqrt{x+1}} \left(x-3 \sqrt{x+1}+4\right)}{8(x+1)^{5/2}}\\f^{(4)}(x)&=\frac{e^{\sqrt{x+1}} \left(x \left(\sqrt{x+1}-6\right)+16\sqrt{x+1}-21\right)}{16(x+1)^{7/2}}\\f^{(5)}(x)&=\frac{e^{\sqrt{x+1}}
   \left(x^2+\left(47-10 \sqrt{x+1}\right) x-115
   \sqrt{x+1}+151\right)}{32(x+1)^{9/2}}\end{align}$$Therefore$$\begin{align*}f(0)&=e\\f'(0)&=\frac e2\\\frac{f'(0)}{2!}&=0\\\frac{f^{(3)}(0)}{3!}&=\frac e{48}\\\frac{f^{(4)}(0)}{4!}&=-\frac{5e}{384}\\\frac{f^{(5)}(0)}{5!}&=\frac{3e}{320}\end{align*}$$So, the first terms of the Maclaurin series of $f(x)$ are$$e+\frac e2x+\frac e{48}x^3-\frac{5e}{384}x^4+\frac{3e}{320}x^5$$and therefore the first terms of the Maclaurin series of $e^{\sqrt{x^2+1}}$ are$$e+\frac e2x^2+\frac e{48}x^6-\frac{5e}{384}x^8+\frac{3e}{320}x^{10}.$$
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