 So let's take a very detailed look at evaluating a difference quotient. So let f of x equals 2x squared plus 1. Let's find and simplify the difference quotient f of x plus h minus f of x over h. So the first thing we need to do is find f of x plus h and f of x. So we're given our definition of f of x. So let's start out by dropping every occurrence of x and replacing it with an empty set of parentheses. And remember that whatever we put in one set of parentheses, we've got to put in all the sets of parentheses. So we want to find f of x plus h. So we want to put an x plus h inside this set of parentheses. So it should also go here. We also want to find f of x. So let's put an x inside the empty set of parentheses. So equals means replaceable. So wherever we see f of x plus h, we can replace it with 2 times x plus h squared plus 1. And wherever we see f of x, we can replace it with 2x squared plus 1. Where because we're subtracting f of x, we should throw the whole thing inside a set of parentheses. And remember arithmetic is bookkeeping and algebra is generalized arithmetic. We had all of this over h and so it's still over h. And so we found our difference quotient, but we do need to simplify it. So right now our numerator is a mess of additions and products and powers and subtractions. Let's see what we can do to simplify this so it has fewer operations. So the first thing we might do is we might expand out this first term to times x plus h squared. So paper is cheap. Let's go off to the side and write down our computations to x plus h squared. Well that exponent means that we have x plus h times x plus h. We need to expand this product. So pick your favorite way of multiplying two polynomials. I like area models. So here we have a rectangle with sides x plus h by x plus h. And we'll find each of these four products x squared, xh, hx, and h squared. So our product is the sum x squared plus xh plus hx plus h squared. Now because multiplication is commutative, xh is the same as hx. So I'll reverse the order of the factors. And so now we have two terms that are the same. So we can simplify this a little bit further. These become 2hx. And algebra is generalized arithmetic and arithmetic is bookkeeping. We had an x squared. We still have an x squared. We had an h squared. We still have an h squared. We had a 2. We still have a 2. But wait, there's still more. I can expand using the distributive property. So that'll be 2 times x squared plus 2 times 2hx plus 2 times h squared. And we might try to simplify this a little bit further. 2x squared. Well, there's not a whole lot we can do with that, so we'll just copy it down. 2 times 2hx. We can multiply these two factors of 2 together to get 4hx. And we can't do much with 2h squared. So we'll just copy it down. Remember all of this was to find 2 times x plus h squared. So equals means replaceable. So instead of writing 2 times x plus h squared, I'll write 2x squared plus 4hx plus 2h squared. And again, that just replaces this. We still have a 1. We still have a minus 2x squared plus 1. And we still have an over h. So let's see if we can simplify the numerator. So when we subtract a polynomial, we subtract every term of that polynomial. So minus 2x squared plus 1. Well, that requires us to subtract 2x squared and to subtract 1. We'll rearrange our terms a little. So we have like terms 2x squared and minus 2x squared. We'll write those next to each other. There's nothing like 4hx. So we'll just copy that down. There's nothing like 2h squared. So we'll copy that. This plus 1 is a constant term, as is the minus 1. So we'll put those together. And we've already copied this minus 2x squared. So we won't copy it a second time. And we can do a little simplification. 2x squared minus 2x squared. Well, that's gone. 4hx still there. 2h squared still there. And plus 1 minus 1 is also gone. And so we can replace our original numerator with 4hx plus 2h squared. So we can try to remove a common factor. But that requires us to actually have a factorization. So let's take a look at this numerator 4hx plus 2h squared. And while we could factor this completely, keep in mind that a factor is only relevant if it's a common factor. And so since my denominator has factor h, well, it is just h, then my only question is, will h factor from the numerator? And we see that both terms do have a factor of h. So I can move that common factor. So 4hx, I can write that as h times 4x. 2h squared, I can write that as h times 2h. And having moved the common factor, I can now remove that common factor h times 4x plus 2h. Equals means replaceable, so our numerator can be replaced with h times 4x plus 2h. And now I have a common factor of h in numerator and denominator. I can remove that common factor, leaving me with 4x plus 2h.