If $f(x)=\sin ax + e^{-ax}$ where $a$ is a constant, what is $f^{(18)}(x)$? I'm a high school senior taking AP Calculus and this question popped up on a test. 
If $f(x) = \sin ax + e^{-ax}$ where $a$ is a constant, what is $f^{(18)}(x)$?
The answer choices were:
(a) $a^{18}(e^{-ax} + \sin ax)$  $\qquad$ (b) $a^{18}(e^{-ax} - \sin ax)$
(c) $a^{18}(e^{-ax}- \cos ax)$ $\qquad$  (d) $a^{18}(\sin ax - e^{ax})$
I had never seen the notation $f^{n}(x)$ before (usually if it's first or second derivative we get $f'(x)$ or $f''(x)$, or if it's greater than that we get it in $d^{n}y/dx^{n}$ form), but I assumed it wanted the $18$th derivative of f(x) (?). Obviously there is a much easier way to do this than writing out $18$ derivatives, especially since there is probably a simple pattern involving the sign and the constant  a because the derivatives of $\sin ax$ and $e^{ax}$ are straightforward. I tried to just think through it in my head, but in a time limit situation that's not a very useful strategy. 
Any help?
 A: Hint. Show that: 
1) the $n$-th derivative of $e^x$ is $e^x$;
2) the $n$-th derivative of $\sin(x)$, is $(-1)^{n/2}\sin(x)$ if $n$ is even and it is
$(-1)^{(n-1)/2}\cos(x)$ if $n$ is odd; 
3) the $n$-th derivative of $f(ax)$ is $a^n\cdot  f^{(n)}(ax)$.
Then you will able to compute the $n$-derivative of $f(x)=\sin ax + e^{-ax}$.
A: $$f(x)=\sin ax +e^{-ax}$$
Let $f(x)=g(x)+q(x)$, where $g(x)=\sin ax, q(x)=e^{-ax}$. Then 
$$f^{(18)}(x)=g^{(18)}(x)+q^{(18)}(x)$$
$q^{(n)}(x)=(-1)^na^ne^{-ax}$
and
$f^{(4k+1)}=a^{4k+1}\cos ax$
$f^{(4k+2)}=-a^{4k+2}\sin ax$
$f^{(4k+3)}=-a^{4k+3}\cos ax$
$f^{(4k)}=a^{4k}\sin ax$
A: It will help to write the first few derivatives of the function $f$:
$$ f_1= a\cos ax-ae^{-ax}$$ $$ f_2=-a^{2}\sin ax+a^{2}e^{-ax}$$ $$ f_3=-a^{3}\cos ax-a^{3}e^{-ax}$$ $$ f_4=a^{4}\sin ax+a^{4}e^{-ax}$$ and so on. From this we can see the general term as: $$f_{4n+1}=a^{4n+1}\cos ax-a^{4n+1}e^{-ax}$$ $$ f_{4n+2}=-a^{4n+2}\sin ax +a^{4n+2}e^{-ax}$$ $$ f_{4n+3}=-a^{4n+3}\cos ax -a^{4n+3}e^{-ax}$$ $$ f_{4n+4}=a^{4n+4}\sin ax +a^{4n+4}e^{-ax}$$ for all $n \in \mathbb N\cup 0$ and $f_n$ is the $n$-th derivative of $f$. Here $f^{18}(x)$ gives us $f_{18} =-a^{18}\sin ax +a^{18}e^{-ax}=a^{18}(e^{-ax}-\sin ax)$. Hence option $(b)$ is correct.
A: Differentiate $\sin (ax)$ twice, you will have $$\frac{d^2\sin(ax)}{dx^2}=-a^2 \sin(ax)$$
Differentiate $\sin (ax)$ four times, you will have $$\frac{d^2\sin(ax)}{dx^2}=a^4 \sin(ax)$$
Think of it as a cycle of $4$; when differentiate it $16$ times, we will have a positive sign; two more differentiation will result in a negative sign. $$\frac{d^{18}\sin(ax)}{dx^{18}}=-a^{18} \sin(ax)$$
Differentiate $e^{-ax}$ is easy. $$\frac{d^{18}e^{-ax}}{dx^{18}}=(-a)^{18} e^{-ax}$$
