If acceleration $a = v\cdot\frac{\mathrm dv}{\mathrm dx}$ , then why isn't acceleration always zero when velocity is zero? We know that,
$$\begin{align}
a&=\frac{\mathrm dv}{\mathrm dt}\\[3pt]
&=\frac{\mathrm dv}{\mathrm dx}\cdot\frac{\mathrm dx}{\mathrm dt}\\[3pt]
&=v\cdot~\frac{\mathrm dv}{\mathrm dx}
\end{align}$$
According to this equation, whenever velocity is $0$, acceleration becomes $0$ as well. 
However, we know that this is not always the case. 
For example, when we throw a ball upwards, at its maximum height, its velocity becomes $0$ but the acceleration is non-zero ($g$ downwards).
Can someone please help me understand what is wrong in my approach? 
 A: It is perhaps worth considering your equation in terms of finite changes rather than infinitesimal changes. 
If $a = \dfrac {\Delta v}{\Delta t}$ then for a given interval of time $\Delta t$ there is a fixed change in the velocity $\Delta v$.  
$a = \dfrac{\Delta v}{\Delta x}\dfrac{\Delta x}{\Delta t}$ 
If the second (velocity) term on the right hand side of the equation for acceleration  is getting smaller and smaller then $\Delta x$ must be must be getting smaller and smaller.  
If $\Delta x$ is getting smaller and smaller then the first term is getting bigger and bigger.  
Since the rates of increase and decrease for the two terms, which are both dependent on $\Delta x$, are the same then the product of the two terms stays constant.  
Note that the first term blows up if you try and make $\Delta x$ equal to zero.
A: Calculus doesn’t work that way. You can’t regroup $(dv/dv)(dv/dt)$ as you’re trying to. 
The basic parts of $dv/dt$ are $v$, the velocity, and $d/dt$, an operator that forms a derivative. 
