Can the independence of cts functions always be checked thru differentiation? I'm wondering about the significance of the differentiation in this excerpt from a linear algebra book.  I guess because its a linear algebra book, they kind of glossed over why they are using the calculus part.  Seems like maybe the rate of change is important to show how they are independent or something?

 A: You can prove the following theorem on your own:

Two differentiable functions $y_1(t)$ and $y_2(t)$ are linear independent if and only if the functions $Y_1(t)=(y_1(t),y_1'(t))$ and $Y_2(t)=(y_2(t),y_2'(t))$ are linear independent. 

Consequently the system is given by
$$\begin{pmatrix}
e^t & e^{2t}\\
e^t & 2e^{2t}
\end{pmatrix}\begin{pmatrix} a\\ b\end{pmatrix}=\begin{pmatrix} 0\\0 \end{pmatrix}$$
which has the unique solution $a=0$ and $b=0$.
A: I think the following question is behind that Example. 

If $(f_1,…,f_n)$ are linear independent in $V$, are $(Tf_1,…,Tf_n)$ linear independent in $V'$, with the linear mapping $T:V→V'$? 

The mapping $T$ represents the differentiation $\frac{d}{dt}$,which is linear, as 
$$\frac{d}{dt}af+bg = a\frac{d}{dt}f+b\frac{d}{dt}g.$$
Now the statement is true, iff $\ker(T)=\{0\}$.    ($⇔ T$ injective) 
Think about constant functions. 
So if you exclude constant functions, which is true, since you look at exponential functions, everything works as intended. 
