I am learning about implicit differentiation and although I understand the rules, I am having hard time understanding the notation.

I have failed to understand differentiation with regards the variables i.e. differentiation of a variable with respect to x and y. Lets start with a simple example. I will carry the calculations up to the point where I could ask all my questions:

(1) 7y

^{4}+x

^{3}y+x=4

(2) d/dx[7y

^{4}+x

^{3}y+x]=d/dx[4]

(3) d/dx[7y

^{4}]+d/dx[x

^{3}y]+d/dx[x]=0

So far so good.

We are at the point where I could ask my first question. Lets isolate the term to the right of the equal sign and assume that it is a separate function y=4 then dy/dx would be 0. What would the derivative be with respect to y? Continuing:

(4) 28y

^{3}[dy/dx]+x

^{3}[dy/dx]+3yx

^{2}+1=0

This is correct … here's why (taking each term separately):

d/dx[7y4] = d/dy [7y^4] * [dy/dx] …. that's the chain rule

=28y^3 * [dy/dx] d/dx [x^3y] … this is a product so you need to use the product rule ...let u= x^3 and v=y

=uv' + vu' (the derivatives are wrt x)

=x^3*d/dx(y) + y*d/dx(x^3)

=x^3 * [dy/dx] + y*3x^2

= x^3 [dy/dx] + 3x^2y

The third term is obvious.
I think here we first assume that y=u and then take du/dx, am I correct?

No But then why do we continue to carry dy/dx? If we isolate d/dx[7y

^{4}] from (3) above, what would be its derivative with regards to y?

In the second term you can assume that u= 7y^4, so you need to find du/dx = du/dy * dy/dx (Chain rule) That's why the dy/dx is still there.
Please, explain as if you were explaining it to a four year old...

Thanks in advance.