Prove that $\{\cos x, \sin x, e^x, e^{-x}\}$ is a linearly independent subset of $C^\infty (\mathbb R)$ I am going to prove $\cos x, \sin x, e^x$ and $e^{-x}$ is a linearly independent subset of $C^\infty (\mathbb R)$, which is smooth functions.
first we have $a\cos x+b\sin x+ce^x+de^{-x}=0$, WTS that $a=b=c=d=0$.
Suppose $c \ne 0$, then for all $x$ in $\mathbb R$, $ce^x$ is sometimes much larger than other 3 terms which is contradiction. So $c=0$.
So we have $a\cos x+b\sin x+de^{-x}=0$, if $d \ne 0$, then by the same logic in $c$, for all $x$ in $\mathbb R$, $x$ is also sometimes much larger than other 2 terms which is contradiction. So $d=0$.
Here it becomes $a\cos x+b\sin x=0$,
when $x=0, a*0+b*1=0$
when $x=\pi/2, a*1+b*0=0$ there is no such situation that $\cos x=\sin x=0$.
So $a=b=0$.
Above is my proof, we have not learned Wronskian or det, so I could only prove it by definition. While since it is going to prove subset of $C^\infty(\mathbb R)$, is there any correction or improvement for the above? Or is there a more clear way to prove this?

"Sometimes much larger" isn't a precise term, although it can be phrased more rigorously. For example, I suggest considering limits $\lim_{x\rightarrow\pm\infty}$. If $$ a \cos x + b\sin x+ c e^x +  d e^{-x} =0$$
for all $x$, then $$ \lim_{x\rightarrow\pm\infty} (a \cos x + b\sin x+ c e^x +  d e^{-x}) = 0$$
and you can use that to show that $c=0=d$. 
