<s>[INST] Determine the acceleration given the position function of a particle 
The position function of a particle in a test laboratory is $s(t) = \frac{10t}
{t^2+3}$.
  Determine the acceleration of the particle of particle after 4 seconds.

The speed $v(t)$ of the particle is the derivative of the position $\frac{ds(t)}{dt}$.
I got $v(t) = \frac{10(-t^2+3)}{(t^2+3)^2}$
For the acceleration $a(t)$ I tried taking the derivative of $v(t)$ but got the wrong answer at $a(4)$.
$$a(t)=\frac{-2t\cdot (t^3+3)^2-10(-t^2+3)\cdot 2(t^2+3)\cdot(2t)}{(t^2+3)^4}$$
 [/INST] Let $S(t)$ be given by
$$s(t)=\frac{10t}{t^2+3}\tag1$$
Differentiating $(1)$ yields
$$\begin{align}
v(t)&=s'(t)\\\\
&=-10\frac{t^2-3}{(t^2+3)^2}\\\\
&=-10\frac{t^2+3-6}{(t^2+3)^2}\\\\
&=-10\frac{1}{t^2+3}+60\frac{1}{(t^2+3)^2}\tag 2
\end{align}$$
Differentiating $(2)$ yields
$$\begin{align}
a(t)&=s''(t)\\\\
&=20\frac{t}{(t^2+3)^2}-240\frac{t}{(t^2+3)^3}\\\\
&=\frac{20t(t^2-9)}{(t^2+3)^3}\tag 3
\end{align}$$
Evaluting $(3)$ at $t=4$, we find that 
$$a(4)=\frac{560}{6859}$$
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