Solve f''(t) for Vector-Valued Function f(t)

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Question:

If f is a vector-valued function defined by f(t)=(e^(-t), cos(t)), find f''(t).
I'm not even quite sure how to start.

Any help would be loved! Thank you!
 
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Ok, let me clarify. I attempted to take the derivative of each separate part (thus, e^-t became -e^(-t) and so on), but I don't know what to do with it.
 
What is the difference between ...

f'(t) vs f''(t) ?

Also, x=e^{-t} and y=\cos t

v(t)=<x=f(t),y=f(t),z=f(t)>

You can take the derivative of each separately.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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