Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. The slope of the curve Y equal F of X at the point X naughty. F of X naughty is the number M equal the limit as H approaches zero of F of X naughty plus H -, F of X naughty all divided by H provided that the limit exists. The tangent line to the curve at P is aligned through P with the slope. If we were thinking about our slope formula being Y 2 -, y one over X 2 -, X one, this formula truly comes from letting H equal X 2 -, X one. So then H plus X1 would equal X2. So when we have F of X 2 -, F of X1 over X2 minus X1, instead of the X2, we would put in that X1 plus H, and instead of the X 2 -, X one in the bottom, we put in what it equals. Because it's a slope of a curve, we want to figure out what's happening at the specific point, which is why we put the limit as H goes to 0. The derivative of a function F at a point X not denoted by F prime of X not is F prime of X not equal. The limit as H goes to 0 of F of X not plus H -, F of X not all over H. Provided this limit exists, The following are all interpretations for the limit of the difference quotient. So we can think of it as the slope of the graph at a single point. We could think of it as the slope of the tangent to the curve at a given point, the rate of change with respect to X at some given point, or the derivative at a point. So if we looked at an example or two, say we have Y equal 4 -, X ^2 and we're given the point -1 three and we want to find the tangent line. So we're going to find the slope first. The limit as H goes to 0 of 4 -, X + H ^2 -, 4 -, X ^2 all over H. Now we could put our numbers in if we wanted to, but we could also do it generically till we get to the end. So the limit as H goes to zero of four minus X ^2 + 2 XH plus H ^2. If I distribute that negative through, we get the -4 + X ^2 at the end. So the limit as H goes to zero 4 -, X ^2 -, 2 XH minus H ^2 -, 4 + X ^2 all over H. So the four -4 are going to cancel the negative X ^2 and positive X ^2 are going to cancel. So we're going to have limit as H goes to 0 of -2 XH minus H ^2 over HI could then think about factoring out an H in the top so that I could reduce. So I'd get negative two X -, H all over H. The limit as H goes to 0 of negative two X -, H is just going to be -2 X. So now if I want the slope at the given point -1 three, so the slope at -1, three is just going to be -2 times my X value of -1 or two. So y - 3 equal 2 * X - -1 If we wanted to put it as a function, we'd get Y equal 2 X +2 + 3 would be +5 if we wanted to find the slope of the curve at the point indicated. Another example, say we have Y equal 1 / X - 1 at X equal 3. The same concept. We're going to have the limit as H approaches 0 of 1 / X + H - 1 - 1 / X - 1 all over H Here we need to get a common denominator. So X - 1 is going to be multiplied by X + H - 1. So the first term has to have an X -, 1 in the numerator and the second one in X + H - 1 in the numerator. And that's all divided by H So now when we continue, the X's are going to cancel X -, 1 -, X -, H + 1, that X + H - 1 * X - 1. This is all being divided by H, which we could really think of being multiplied by 1 / H So we're going to get an H in the denominator, an X + H - 1 and X -, 1. The XS in the numerator cancel the negative ones in the OR the -1 and +1 cancel and we get negative H So this simplifies down to limit as H goes to 0 of -1 / X + H - 1 X -1. When I put H is 0 in I get -1 / X - 1 ^2. So if I want the slope when X is 3, if we thought about X being three, we could figure out YY would be 1 / 3 - 1, so Y is 1 / 3 - 1 or 1/2. So the slope at the .31 half -1 / 3 - 1 ^2 -1 / 4. So y - 1/2 equal negative 1/4 X - 3 Y equal -1 fourth X + 3/4 + 1/2 Y equal -1 fourth X + 5 halves. Thank you and have a wonderful day.