How is $ \frac{dv}{ dt} = a $? I know how , in the physical sense - 
$$\frac {dv}{dt} = a$$
But, even after thinking a lot, I am not able to see the fault in this - 
$$\frac {dv}{dt} = \frac {d(st^{-1})}{dt}
   = \frac {sd(t^{-1})}{dt}
  = s*(-1)*t^{-2}
 = \frac {-s}{t^2}
  = \frac {-v}{t}
 =  -a$$
I know something is terribly wrong here but I'm just not able to figure out what or where. 
Please keep in mind I'm just a curious 16 year old. Any help would be greatly appreciated. 
 A: 
I know how , in the physical sense -  $$\frac {dv}{dt} = a$$
But, even after thinking a lot, I am not able to see the fault in this
  -  $$\frac {dv}{dt} = \frac {d(st^{-1})}{dt}    = \frac {sd(t^{-1})}{dt}   = s*(-1)*t^{-2}  = \frac {-s}{t^2}   = \frac
> {-v}{t}  =  -a$$
I know something is terribly wrong here but I'm just not able to
  figure out what or where.  Please keep in mind I'm just a curious 16
  year old. Any help would be greatly appreciated.

Here's an example. Suppose an object is traveling at a constant velocity. Then
$$
s=v_0t
$$
where $v_0$ is constant in time. I.e., we know that $dv_0/dt=0$ and $ds/dt=v_0$.
So, for this example, you might get confused if you rearrange and write
$$
v_0 = \frac{s}{t}
$$
and then the RHS looks like it might not be constant... but it has to be, so what gives? 
Well, in general:
$$
\frac{d}{dt} \left( \frac{s}{t} \right)
=-\frac{s}{t^2}+\frac{1}{t}\frac{ds}{dt}
$$
because I have to differentiate both terms: the $s$ term; the $\frac{1}{t}$ term. This is just an application of the product rule of differentiation. The above result holds in general.
And, for the example case of constant velocity, plugging in $s=v_0t$, the above equation becomes:
$$
-\frac{s}{t^2}+\frac{1}{t}\frac{ds}{dt}=-\frac{v_0t}{t^2}+\frac{1}{t}v_0=0
$$
So, the problem you are having is that you have only differentiated one of the terms, the $1/t$ term, and not the other, the $s$ term.
