<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 have to compute the derivatives
$f(x)=e^{\sqrt{x^2+1}}$
$f^1(x)=\dfrac{e^{\sqrt{x^2+1}} x}{\sqrt{x^2+1}},\\f^2(x)=\dfrac{e^{\sqrt{x^2+1}} \left(\sqrt{x^2+1} x^2+1\right)}{\left(x^2+1\right)^{3/2}},\\f^3(x)=\dfrac{e^{\sqrt{x^2+1}} x \left(x^4+x^2+3 \sqrt{x^2+1}-3\right)}{\left(x^2+1\right)^{5/2}},\\f^4(x)=\dfrac{e^{\sqrt{x^2+1}} \left(6 \left(3-2 \sqrt{x^2+1}\right) x^2+3 \left(\sqrt{x^2+1}-1\right)+\sqrt{x^2+1} x^6+\left(\sqrt{x^2+1}+6\right) x^4\right)}{\left(x^2+1\right)^{7/2}},\ldots$
And then evaluate them at $x=0$
$f^0(0)=e,\;f^1(0)=0,\;f^2(0)=e,\;f^3(0)=0,\;f^4(0)=0,\;f^5(0)=0,\;f^6(0)=15 e,\;f^7(0)=0,\;f^8(0)=-525 e,\;f^9(0)=0,\;f^{10}(0)=34020 e$
And use the formula 
$$f(x)=\sum _{n=0}^{\infty } \frac{f^n(0) x^n}{n!}$$
and get
$$f(x)=e\left(1+\frac{x^2}{2} +\frac{x^6}{48}-\frac{5 x^8}{384}+\frac{3 x^{10}}{320}+O(x^{11})\right)$$
Hope this helps
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