In Example 2, everything was fine, what I was doing matched perfectly with the book until the part "Setting this equal to zero gives *equation*". How is this? The expression on the left has no p, how can it have a p on the right? If the derivative dS/dx is set to 0 how is an x on the right, at the other side of =? I understand the rest. It's just that I don't see how one expression results in the other. That's all. Thank you.

 

P.S: Sorry if images are annoying, english is not my first language, so this way it's easier for me to explain and for you to understand.

 

4 hours ago, KFS said:

How is this?

One part of the equation is moved to the right, note that the sign is changed (x-p) becomes (p-x) and the zero is not explicitly displayed. I highlighted that in image* below.

The derivative dS/dx is set to 0, see the first zero i added. The expression for dS/dX is then expressed as function of x, p, q and restructured. On the second row I added the zero again to show the equality. 

 

*) Had to use phone, hard to create a good picture.

 

6 hours ago, Ghideon said:

 

One part of the equation is moved to the right, note that the sign is changed (x-p) becomes (p-x) and the zero is not explicitly displayed. I highlighted that in image* below.

The derivative dS/dx is set to 0, see the first zero i added. The expression for dS/dX is then expressed as function of x, p, q and restructured. On the second row I added the zero again to show the equality. 

 

*) Had to use phone, hard to create a good picture.

 

Thanks, that's been helpful.

10 minutes ago, KFS said:

Thanks, that's been helpful.

Thanks for the feedback!

It's always good to know that an explanation was helpful so similar cases can be addressed in similar ways in the future. 
(Or that an explanation was not helpful and improvements are needed)