Gradient of kinetic energy I have got very silly doubt. We know we can write force as:
$$\mathrm{Force}=-\frac{dU}{dx}$$
So why we can't write kinetic energy as
$$\mathrm{Force}=\frac{dK}{dx}$$
even though we get
$$\int F\cdot dx = K_f - K_i$$
 A: It all comes to convenience.
We can write $F=\frac{dK}{dx}$, but kinetic energy often is not wrote as a function of $x$.
Potential energy often is written as a function of $x$. Like gravitational potential energy is written as $U(x)=-\frac{Gm_1m_2}{x}$. Force often is also written as a function of $x$ particularly field forces (like gravity: $F(x)=\frac{Gm_1m_2}{x^2}$), which vary over the distance from origin of reference frame.
That's why: $$F(x)=\frac{U(x)}{dx}$$
, is more convenient to write.
However if we have kinetic energy as a function of $x$, then we can apply $F(x)=\frac{dK(x)}{dx}$
A: 
We know we can write force as: $$\mathrm{Force}=-\frac{dU}{dx}$$

Actually, this is a little incorrect. The correct relationship is $$\vec F = - \nabla U = \frac{\partial U}{\partial x} \hat x + \frac{\partial U}{\partial y} \hat y + \frac{\partial U}{\partial z} \hat z$$ where $\hat x$ is the unit vector in the x direction. Remember force is a vector and the potential is a scalar. Of course, if you are doing a "motion on a straight line" problem you can neglect the vector nature of force.
However, even if you are doing a unidimensional problem, you still need to respect the difference between a partial derivative and a total derivative. In this sense, we have $$KE = \frac{1}{2}mv^2$$ This does not explicitly depend on $x$, it only explicitly depends on $v$, the derivative of $x$. So $$\frac{\partial}{\partial x}KE =  \frac{\partial}{\partial x} \frac{1}{2}mv^2 = 0$$ which is not particularly useful as far as I can tell.
A: Taking into account Newton's 3rd Law, a force is really an interaction between two particles, where each particle exerts the same magnitude force on the other, in opposite directions. Therefore, there can't be a force if there's only one particle!
Conversely, kinetic energy is a property of a single particle$^1$. Therefore, it cannot be used to define a force exerted on the particle.
Thinking about things carefully, we can see potential energy and force as two alternative ways of describing the same phenomenon.  This can't be done with kinetic energy and force, because a force causes a change in kinetic energy, as quantified through the work-energy principle.
--
One of the lines of reasoning that the OP used to think that perhaps $F = dK/dx$ might make sense is that
$$
\Delta K = \int_C\vec{F}\cdot d\vec{r}\,,
$$
and so, since the kinetic energy is the integral of the force, the force should be the derivative of the kinetic energy.  However, the force appearing in this expression is the net force acting on the object, and so if a relationship like $F = dK/dx$ was to hold, $F$ would have to refer to the net force acting and not an individual force exerted by, say, a gravitational field.
Furthermore, there's no such thing as potential energy to begin with if we don't define it as the derivative of the force (at least, if we are starting with forces as the fundamental notion).
--
$^1$....relative to some reference frame, which actually does require other particles to exist in order for $K$ to be well defined, so there's a bit of sophistry here.
