 We can find difference quotients for other functions as well, so for example, we might take a look at a rational function. How about f of x equals 1 over 2x plus 5, and this time we'll take a look at the same difference quotient. Again, as a good starting point, let's go ahead and find expressions for f of x and f of x plus h. Again, f of x is easy, that's just what the function is. For f of x plus h, again, strongly advise you to drop the value of x entirely, it's really just a placeholder, replacing every occurrence of x with a set of empty parentheses, set of empty parentheses there, and whatever goes in the 1, and we would like to put x plus h in this one, should also go in all of the other sets of parentheses. So f of x plus h, 1 over 2 times x plus h plus 5, and we'll do a little bit of algebra to make our lives easier. We'll expand out that 2 times x plus h. So there's our expression for f of x plus h, our expression for f of x, and now we're ready to set down the difference quotient. So how does that work? Well, I want to have f of x plus h minus f of x all divided by h, so that's going to look like this. So here's my f of x plus h minus f of x all over h, and I have this rather horribly complicated rational expression over on the right-hand side, but I can get rid of the worst part of the fractions by multiplying through by the common denominator of all fractions present. So here my common denominator is the product of the two denominators, 2x plus h, 2h plus 5, and 2x plus 5. So if I multiply everything by that product of these two denominators, divided by the product of the same two denominators, because I don't want to actually change the value of the expression, I'm going to be able to do some considerable amount of simplification. So I'll expand that out. So I have this mess times the first fraction, this mess times the second fraction, and then all subtracted. And then the denominator h times this times this. At this point it's probably not worth multiplying that out, because it's a lot of work and we hope that something nice will happen. And it turns out something does. So what happens? Well, this 2x plus 2h plus 5 in the denominator cancels with this 2x plus 2h plus 5 in the numerator, and so our first expression simplifies. Likewise, the 2x plus 5 here in the denominator cancels with the 2x plus 5 here in the factor, so our second term also simplifies. So the first term after the two common factors cancel, 2x plus 5. Our second term after the two common factors cancel, what's left over? 2x plus 2h plus 5. Still in parentheses, because parentheses are cheap, and let's go ahead and expand that out. That's 2x plus 5 minus everything minus 2x minus 2h minus 5, and we do have a couple of things that'll cancel. Here's a 2x minus 2x of 5 minus of 5, and those things will drop out. Minus 2h, and we have one more thing that can cancel. There's an h up top, there's an h on the bottom, and again the key qualifier we should add here is we can cancel those h's out as long as h itself is not equal to 0. So we'll do the cancellation, but then we'll throw in some fine print that says h can't be equal to 0. And there is our difference quotients. Thank you.