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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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