Electric Field at a Point on the x-Axis Due to Charges

[gracy]

Homework Statement

The potential at a point x(measured in µm)due to some charges situated on the x-axis is given by ##V(x)##=##\frac{20}{x^2-4}##V.The electric field ##E## at x=4µm is ?

Homework Equations

##E##=-##\frac{∂V}{∂r}##

The Attempt at a Solution

Actually I have solution but still I am unable to understand.So ,instead of my attempt at a solution .I 'll post solution itself.
##Ex##=-##\frac{∂V}{∂x}## (1)

= -##\frac{d}{dx}(\frac{20}{x^2-4})## (2)

=##\frac{40x}{(x^2-4)^2}## (3)

##Ex## at x=4µm =##\frac{10}{9}##V/µm
Here I am not able to understand how it proceeded from (2) to (3)

 

[blue_leaf77]

Look at the second equation in https://en.wikipedia.org/wiki/Quotient_rule.

 

[gracy]

And how to proceed from (3) to answer?

 

[blue_leaf77]

You just plug in x=4 into the expression, right?

 

[gracy]

right?

Yes,of course.