McLaurin expansion for $f(x)=e^{\sin{x}}$ of the 4:th order. By the 4:th order, they mean using the 4:th derivative. But the differentiation gets a bit ugly quite fast, so instead of computing all the derivatives , I should be able to use the standard expansions:
$$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+B(t) \\ \sin{x}=x-\frac{x^3}{3!}+B(x),$$
Where $B(t),B(x)$ are functions that are bounded close to $x=0.$ Correct me if I'm wrong, but both of the expansions above are of 4:th order since for the sine function we have two terms vanishing because they become zero?
I'm not sure how I should fuse the separate expansions of $e^t$ and $\sin{x}$ to one expansion. Setting $t=\sin{x}$ seems to give me an expression tedious tedious to work with.
 A: You already have
$$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+O(t^5)$$
and
$$\sin{x}=x-\frac{x^3}{3!}+O(x^5),$$
So just substitute:
$$e^{\sin x} = 1 + \left(x-\frac{x^3}{3!}+O(x^5)\right) + \frac{1}{2!}\left(x-\frac{x^3}{3!}+O(x^5)\right)^2 \\+ \frac{1}{3!}\left(x-\frac{x^3}{3!}+O(x^5)\right)^3 +\frac{1}{4!}\left(x-\frac{x^3}{3!}+O(x^5)\right)^4 + O\left(\left(x-\frac{x^3}{3!}+O(x^5)\right)^5\right)\\
=1 + \left(x-\frac{x^3}{3!}\right) + \frac{1}{2!}\left(x^2-2\frac{x^4}{3!}\right) + \frac{1}{3!}\left(x^3\right) +\frac{1}{4!}\left(x^4\right) + O\left(x^5\right).$$
Gather like terms and bob's your uncle.
$$e^{\sin x} = 1 + x + \frac{x^2}{2!}  -\frac{x^3}{3!}+\frac{x^3}{3!} -2\frac{1}{2}\frac{x^4}{3!}+\frac{x^4}{4!}+O(x^5)\\=1 + x + \frac{x^2}{2!} - \frac{x^4}{8}+O(x^5).$$
A: Yet another way:
$$ e^{\sin x} = e^{x}\cdot e^{-x^3/6}\cdot e^{o(x^4)} =\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\right)\left(1-\frac{x^3}{6}\right)+o(x^4)$$
and the computations are now straightforward.
A: ziggurism's answer is the most straightforward (using substitution), and you must master it. But here's another way: let $f=\exp\circ\sin$. Then $f'=f\times\cos$. Let $a,b,c,\ldots$ be the coefficients in the Taylor–Young expansion of $f$ at $0$:
$$f(x)\underset{x\to0}=a+bx+cx^2+dx^3+ex^4+o\bigl(x^4\bigr).$$
By the way, we know that $a=1$.
We also have (since $f$ is of class $C^\infty$),
$$f'(x)\underset{x\to0}=b+2cx+3dx^2+4ex^3+o\bigl(x^3\bigr).$$
Now, writing $f\times\cos=f'$
$$\Bigl(1+bx+cx^2+dx^3+o\bigl(x^3\bigr)\Bigr)\left(1-\frac{x^2}2+o\bigl(x^3\bigr)\right)=b+2cx+3dx^2+4ex^3+o\bigl(x^3\bigr)$$
and expanding and identifying the terms with same degree (using the uniqueness of a Taylor–Young expansion) yields:


*

*Constant terms: $b=1$,

*Terms of degree $1$: $b=2c$, hence $c=1/2$,

*Terms of degree $2$: $c-1/2=3d$, hence $d=0$,

*Terms of degree $3$: $d-b/2=4e$, hence $e=-1/8$.


Conclusion:
$$\mathrm{e}^{\sin(x)}\underset{x\to0}=1+x+\frac{x^2}2-\frac{x^4}8+o\bigl(x^4\bigr).$$
A: There is an easy way to find derivatives of $f(x)$ at $x=0$. 
Notice that $f'(x)= f(x)\cos(x)$ therefore $f'(0)=1$. So $f''(x) = f'(x)\cos(x)-f(x)\sin(x)$ therefore $f''(0)=1$. Then $f'''(x) = f''(x)\cos(x)-2f'(x)\sin(x)-f(x)\cos(x)$ therefore $f'''(0)=0$. And $f^{(4)}(x)=f'''(x)\cos(x)-3f''(x)\sin(x)-3f'(x) \cos(x)+ f(x)\sin(x)$. Therefore $f^{(4)}(0)=-3$. We get $f(x) = x + (x^2)/2 - (x^4)/8+ \dotsb$
