Derivative of dot product vs derivative of scalars Suppose $\vec{v}(t)$ is the velocity (vector) function. Then:
$$\frac{\mathrm{d} (\vec{v}\cdot\vec{v})}{\mathrm{d} t}=2 \vec{v} \cdot \frac{\mathrm{d} \vec{v} }{\mathrm{d} t}=2 \vec{v} \cdot \vec{a}=2va \cos \varphi$$
where $\vec{a}$ is the acceleration vector and $\varphi$ - the angle between $\vec{a}$ and $\vec{v}$.
On the other hand:
$$\frac{\mathrm{d} (v\cdot v)}{\mathrm{d} t}=2 v \cdot \frac{\mathrm{d} v }{\mathrm{d} t}=2 v a \neq \frac{\mathrm{d} (\vec{v}\cdot\vec{v})}{\mathrm{d} t}$$
Although $\vec{v}\cdot\vec{v}=v\cdot v$ . Where is my mistake? I ask this because often times in physics I see the substitution $\vec{v} \cdot \vec{v}=v^2$ used in differentiation, although the results we get are different.
 A: You can say that $\vec v\cdot \vec v = v\cdot v$ and for the same reason, you can also say that $$\vec v\frac{d\vec v}{dt}= v\frac{dv}{dt}$$
Also you are right that $v\frac{dv}{dt}=va$ that is, $\frac{dv}{dt}=a$ .
Your only mistake is that while performing the below operation,

$$2\vec v \cdot \vec a =2va\cos \psi$$

Actually $2\vec v \cdot \vec a \not =2va\cos \psi$. 
The correct step is $2\vec v \cdot \vec a =2|\vec v||\vec a|\cos \psi$
And $|\vec a| \not= a$ since 
$$\vec v=v_x \hat i+v_y \hat j+v_z \hat k$$
$$d\vec v=dv_x \hat i+dv_y \hat j+dv_z \hat k$$
$$\vec a=\frac{d\vec v}{dt}=\frac{dv_x}{dt} \hat i+\frac{dv_y}{dt} \hat j+\frac{dv_z}{dt} \hat k$$ 
Hence $$|\vec a|=\sqrt{(\frac{dv_x}{dt})^2+(\frac{dv_y}{dt})^2+(\frac{dv_z}{dt})^2}$$
But $$v=|\vec v|=\sqrt{(v_x)^2+(v_y)^2+(v_z)^2}$$
and $$a = \frac{dv}{dt} = \frac{1}{v}\cdot \left(v_x\frac{dv_x}{dt}+v_y\frac{dv_y}{dt}+v_z\frac{dv_z}{dt}\right) \not = |\vec a|$$
For further clarification, you can refer to this almost similar question of mine on PhySE.
A: \begin{eqnarray*}
\frac{d\mathbf{v}}{dt} &=&\mathbf{a} \\
\frac{dv}{dt} &=&\sum_{j}\frac{dv}{dv_{j}}\frac{dv_{j}}{dt}=\sum_{j}\frac{%
v_{j}}{v}a_{j}=\frac{1}{v}\mathbf{v\cdot a} \\
v\frac{dv}{dt} &=&\mathbf{v\cdot a}
\end{eqnarray*}
A: The first equation is the more general and correct one.
But your second equation is false.
Do not forget that $v = \|\vec v\| = \sqrt{\vec v.\vec v} = \sqrt{\sum_i v_i^2}$ in $L_2$-norm.
$\frac{d(v^2)}{dt} = 2v \frac{dv}{dt}$ (A multiplication between two scalars not a scalar product between two vectors)
The derivation of the second term in the right-hand side (that of a norm that is to say a square root) gives:
$\frac{dv}{dt} = \frac{d\sqrt{\vec v.\vec v}}{dt}=\frac{\frac{d(\vec v.\vec v)}{dt}}{2\sqrt{\vec v.\vec v}}=\frac{2\vec v.\vec a}{2v}=\frac{\vec v.\vec a}{v}$   
Finally by multiplying this term by $2v$ you get the expected result:
$\frac{d(v^2)}{dt} = 2v \frac{\vec v.\vec a}{v} = 2\vec v.\vec a$
