Force acting on a dipole placed in a non-uniform electric field For an electric dipole is placed in a non-uniform electric field, why does the net force act in the direction of increasing electric field?
 A: This is best understood by approximating the dipole as a pair of finite charges $\pm q$ separated by a finite distance $d$. In a uniform electric field, the electrostatic forces on each of the charges will cancel out exactly, but in a non-uniform one the forces on the two will be slightly different, leading to a slight imbalance and therefore  a non-zero net force. As you take the distance to zero, the difference in electric field goes to zero, but the charge also grows to exactly cancel it out.
To be more quantitative, suppose the negative charge is at $\mathbf r$ and the positive charge at $\mathbf r+d\mathbf n$. The total force is then
$$
\mathbf F=q\left[\mathbf E(\mathbf r+d\mathbf n)-\mathbf E(\mathbf r)\right].
$$
To get the correct form for the limit, change from the charge $q$ to the electric dipole $p=qd$, to get
$$
\mathbf F=p\frac{\mathbf E(\mathbf r+d\mathbf n)-\mathbf E(\mathbf r)}{d}.
$$
The true force on a point dipole is the limit of this as $d\to0$,
$$
\mathbf F=p\lim_{d\to0}\frac{\mathbf E(\mathbf r+d\mathbf n)-\mathbf E(\mathbf r)}{d},
$$
and this is exactly the directional derivative along $\mathbf n$, typically denoted $\mathbf n\cdot \nabla$, so
$$
\mathbf F=p\mathbf n\cdot \nabla\mathbf E=\mathbf p\cdot \nabla\mathbf E.
$$
A: The premise of your question is faulty in that when a dipole is placed in a non-uniform electric field, the direction of the net force experienced by it is not always along the direction of increasing electric field. 
(I will work with the assumption that you don't speak of a case in which the dipole is titled with respect to the electric field) Two cases exist--the dipole is either parallel or anti-parallel to the direction of increasing electric field. (The direction of a dipole is in the direction from negative charge to positive charge) 
Let us consider a non-uniform electric field $\vec{E}$ increasing in the direction left to right. 
Case 1: Parallel--

As seen above, the dipole-with dipole moment $\vec{p}$-is aligned parallel to the direction of increasing non-uniform electric field, $\vec{E}$.
The direction of force experienced by a positive charge due to an electric field is along the direction of the field while that by a negative charge is along the opposite direction to the field. In this case, since the magnitude of non-uniform electric field increases from left to right, the force experienced by charge $+$q due to the field will be greater than that experienced by charge $-$q.(The force experienced by a charge $+$Q due to an electric field of magnitude $E$ is given by $F=QE$)
Hence, as shown below, the direction of net force experienced by the dipole is along the direction of increasing electric field--

(The length of the vector represents its magnitude)
Case 2:Anti-parallel--

As seen above, the dipole-with dipole moment $\vec{p}$-is aligned in the opposite direction to that of increasing electric field.
Since the magnitude of electric field increases from left to right, the force experienced by charge $-$q due to the field will be greater than that experienced by charge $+$q. 
Hence, as shown below, the direction of net force experienced by the dipole is opposite to the direction of increasing electric field--

I hope that this clears your confusion.  
A: The net force on dipole in non-uniform electric field is
$$F=P \frac{\mathrm dE}{\mathrm dx}$$
where $P$ is dipole moment and $\frac{\mathrm dE}{\mathrm dx}$ is the change in Electric field w.r.t. to position.
