differentiate to find velocity I need to differentiate $x(t)=e^{2t}(4t^2-3t+1)$  to find velocity and acceleration at $3$ seconds.
I need to use the product rule.
I know $e^{2t} = u$ so $du(t)/dt = 2e^{2t}$ 
and $(4t^2-3t+1) = v$ and $dv= (8t-3)$
so $dy/dx = udv + vdu = (e^{2t})(8t-3)+(4t^2-3t+1)(2e^{2t})$
I am completely stuck where to go next
please help
 A: You've done fine, now just factor out the factor $e^{2t}$ from each term and simplify! (It will make finding the second derivative, acceleration, easier if you simplify).
$$f'(t) = v(t) =(e^{2t})(8t-3)+(4t^2-3t+1)(2e^2t) = (e^{2t})\Big((8t-3)+(8t^2-6t+2)\Big) = e^{2t}(8t^2 + 2t - 1)$$ 
Evaluate at $t = 3$ to obtain velocity:

$$v(3) = f'(3) = e^{6}(8(3)^2 + 2(3) - 1) = e^6(72 + 8 - 1) = 79e^6$$

Then, differentiate $f'(t)$ using the product rule again, and then evaluate at $t = 3$ to find acceleration at time = 3 seconds.

 $$f''(t) = a(t) = 2e^{2t}(8t^2 + 2t - 1) + 4e^{2t}(16t + 2)= 2e^{2t}\Big(8t^2 + 2t - 1 +32t + 4\Big) = 2e^{2t}(8t^2 + 34t + 3).\\$$ $$f''(3) = a(3) = 2e^6(72 + 102 + 3) = 354e^6$$

A: You're confusing yourself with substitution. You're introducing a lot of symbols that might be obscuring the point. Try this:
$$x(t) = \underbrace{e^{2t}}_{\textrm{first term}} \underbrace{\left(4t^2-3t+1\right)}_{\textrm{second term}}.$$
$$\begin{align*}
v(t) = x'(t) &= \left[\frac{d}{dt}(\textrm{first term})\right] \cdot (\textrm{second term}) + (\textrm{first term})\cdot \left[\frac{d}{dt}(\textrm{second term})\right] \\
 &= 2e^{2t}(4t^2-3t+1)+e^{2t}(8t-3).
\end{align*}$$
Do the same thing again to get the second derivative, $a(t) = v'(t) = x''(t)$.
Then, the velocity at $t=3$ seconds is just $v(t=3)$.
