Test linear independence of a set using derivatives 

Is the set $\{1, e^x+e^{-x}, e^x-e^{-x}\}$ linearly independent.


Question: Can I use derivatives to solve this? I'm thinking something like this:
In order for the set to be independent then:
$r + s(e^{x}+e^{-x})+t(e^{x}-e^{-x}) = 0$ if and only if $r$, $s$, and $t$ are equal to $0$.
So,
$r + s(e^{x}+e^{-x})+t(e^{x}-e^{-x}) = 0$
$r + se^{x}+se^{-x}+te^{x}-te^{-x} = 0$ 
$r + (s+t)e^{x}+(s-t)e^{-x} = 0$ 
Here I take the derivative and:
$(s+t)e^{x}-(s-t)e^{-x} = 0$ 
Then, adding last two equations
r+ 2(s-t) = 0
Which shows that the set is not linearly indepent (if I take same values for s and t then equation is 0 (but not s nor t are 0).
Is this reasoning right or am I way two wrong. I'm trying to self-study linear algebra but this is pretty damn hard!
 A: No, you can not. But you can reason as follows:
$$
\begin{align*}
r + s (e^x + e^{-x}) + t (e^x - e^{-x}) &= 0 \\
r + (s + t) e^x + (s - t) e^{-x}        &= 0
\end{align*}
$$
Now, for $x = 0$ you have $r + 2 s = 0$. Try again with $x = 1$ and $x = -1$
(i.e., three easy values), and you get three equations in the three unknowns $r$, $s$ and $t$. You'll find that the only solution is $r = s = t = 0$, i.e., they are linearly independent.
A: For $\,a,b,c\in\Bbb R\,$:
$$a+b(e^x+e^{-x})+c(e^x-e^{-x})=0\iff a+e^x(b+c)+e^{-x}(b-c)=0$$
Choosing $\,x=0\,$ we get
$$a+b+c+b-c=0\iff a+2b=0$$
Choosing $\,x=1\,$:
$$\text{I}\;\;\;\;\;\;a+e(b+c)+e^{-1}(b-c)=0$$
Choosing $\,x=-1\,$:
$$\text{II}\;\;\;\;\;\;a+e^{-1}(b+c)+e(b-c)=0$$
Substract II from I:
$$2c(e-e^{-1})=0\Longrightarrow c=0$$
Try to take it from and show that also $\,a=b=0\,$ and thus the three functions are line. independent.
A: Yes, derivatives can be a very useful tool to test for linear independence. For some details, you may want to read about the Wronskian.
Let's use differentiation efficiently to deal with your functions. 
Suppose that $f(x)=r+s(e^x+e^{-x})+t(e^x-e^{-x})\equiv 0$. Differentiate. We get $f'(x)= s(e^x-e^{-x})+t(e^x+e^{-x})\equiv 0$. Set $x=0$. We get $2t=0$, and therefore $t=0$. 
Differentiate again, and set $x=0$. We get $s=0$. Finally, since $s=0$ and $t=0$, by setting $x=0$ in the equation $f(x)=0$, we get $r=0$.  
