Establish the relation between electric field and potential gradient.

Hint: You could consider a unit positive charge producing electric field around it. Then you could begin from the work done to move it by an infinitesimal distance. Now, you could make it in terms of the electric field produced by the charge. And finally you could derive the required relation.

Complete answer:
Let us begin by considering a unit positive charge producing an electric field E at a certain point containing unit positive charge. Let us move this charge by a distance dx without being accelerated. We should note that the external force that is being applied is equal in magnitude but opposite in direction to the electric field force. That is, we see that the external force and displacement are in opposite directions.
We know that the work done is given by the product of force and displacement, so, here the work done by the external force could be given by,
$dW=-Fdx$…………………………………………… (1)
We know that electric field could be given by,
$E=\dfrac{F}{q}$……………………………………………. (2)
For unit charge equation (2) becomes,
$E=F$………………………………………….. (3)
Now we could substitute (3) in (1) to get,
$dW=-Edx$……………………………………. (4)
We know that electric potential due to a point charge could be defined as the work done so as to move it against an electric field by an infinitesimal distance without being accelerated.
Now equation (4) becomes,
$dV=-Edx$
$\therefore E=-\dfrac{dV}{dx}$
Therefore, we have derived the relation between electric field and electric potential gradient.

Note:
You may wonder why in the definition of electric potential, we stress on the phrase ‘charge is moved without being accelerated’ significant. This makes it possible for the external force being equated in magnitude with force exerted by the electric field. Hence, it is found to be worthwhile to stress on this phrase.