Find acceleration given velocity with respect to distance A particle moves with the velocity given by: 
$$v(s(t)) = \frac{3s(t) + 4}{2s(t)+1}$$
where s(t) is the distance traveled. 
Find the acceleration when s(t) = 2. 
My attempt:
$$a(s(t)) = \frac{\mathrm d}{\mathrm dV(s(t))}\left(\frac{3s(t) + 4}{2s(t)+1}\right) = \frac{(3s(t)+4)2 - 3(2s(t)+1)}{(2s(t)+1)^2}$$
$$a(2) = \frac{10\times 2-3\times 5}{25} = \frac{5}{25} = \frac{1}{5}.$$
The answer is however supposed to be $-\frac{2}{5}$. Why? 
 A: The method is right, but you are wrong about your derivate. 
You have 
$$\frac{\mathrm d}{\mathrm d t}\left(\frac fg\right)=\frac{f'g-g'f}{g^2}$$
and not
$$\frac{\mathrm d}{\mathrm d t}\left(\frac fg\right)=\frac{fg'-f'g}{g^2}.$$
So the answer would be
$$-\frac 15.$$
A: you have $$v = \frac{3s+4}{2s+1}= \frac 32 + \frac 5{2(2s+1)}, s = 2, v = 2.$$differencing you get $$ dv = -\frac{5ds}{(2s+1)^2}=-\frac{ds}{5} \mbox{ at }s = 2.$$ dividing by $dt$ and using $\frac {ds}{dt} = v = 2,$ you get $\frac{dv}{dt} = -\frac25 \mbox{ at } s = 2, v = 2. $
A: You have a function $v$ that gives the velocity $ds/dt$ in terms of position and an unknown function $s$ that gives position in terms of time. So the function that you are differentiating is really $v\circ s$. By the chain rule, $${dv\over dt}(t)={dv\over ds}(s(t))\cdot{ds\over dt}(t)={dv\over ds}(s(t))\cdot(v\circ s)(t)={3(2s(t)+1)-2(3s(t)+4)\over(2s(t)+1)^2}\cdot{3s(t)+4\over2s(t)+1}.$$ Plugging in the known value of $2$ for $s(t)$ produces $-\frac25$ as required.
