 Consider the function f of x equals x squared. What I first wanna do is compute the derivative of x squared and we're gonna do this from the definition. So remind ourselves what the definition of the derivative is. The derivative is the function f prime of x, which for any value x, it's gonna be the limit as h approaches zero of f of x plus h minus f of x all over h. That's one version of the derivative. Another version that more resembles the slope formula is you take the limit as say b approaches x of, well, f of b minus f of x all over b minus x. And so either one of these versions is perfectly fine. In this video, I'm gonna approach this one right here, this more traditional slope approach. Typically speaking, I would say this one with the h is a little bit easier. But again, for the sake of this video, I'm gonna approach the second one there. So if we use the specific function f in this situation, our derivative f prime of x is gonna look like the limit as b approaches x. We're gonna get f of b, which is gonna be b squared minus f of x, which is x squared. And this sits above b minus x. So like I said, if we just plug in x equals b, we're gonna get zero over zero. Not much we can do with that except for the hope that there is something better we can do. Instead, what we need to do here is we're going to factor the numerator. The reason that we get zero over zero is because these two polynomials, b squared minus x squared and b minus x have a common factor. The numerator is actually a difference of squares. We get b minus x and we get b plus x on the top. Then there's a b minus x in the denominator. We see that the factor of b minus x on top cancels with the factor of b minus x in the bottom. And so this ratio would simplify just to be b plus x. Again, as b approaches x right here. Now that we've removed the discontinuity here, we can allow b to become x. So we end up with x plus x, which is equal to two x. So this is the derivative of x squared. The derivative of x squared is two x. Well, what does this two x actually measure then? So recall our function f of x is itself x squared, the parabola, the standard parabola, which you see on the screen illustrated here in yellow. We discovered that its derivative, f prime of x is equal to two x. So what this is telling us is that if we pick any coordinate in the domain, any x coordinate in the domain of the function and we look at the associated tangent line at that point, the slope of the tangent line will just be two times the x coordinate. That's what we're trying to measure here. And so let's look at some specific examples. Let's calculate f prime of three and find out what the tangent line of f at x equals three would be. Well, now that we've already computed the derivative, computing f prime of three is gonna be a cinch, right? Cause we just have to take f prime of three, which by the formula is two times three, which is equal to six. This tells us that the slope of the tangent line at x equals three is gonna be six. So in general, a tangent line is gonna look like y minus f of the value, in this case, it's three. Then it's gonna equal f prime of that value and again, three times y times x minus three right here. So in general, your slope intercept form or your slope point point, excuse me, y minus y one equals m times x minus x one, where x one, y one are a specific point on the line for the line, the tangent line, the point of tangency is the only point we know on the line right now. And then the slope is the derivative, the slope of the tangent line is the evaluation of the derivative because the derivative gives you the formula for the tangent line here. And so plug in the values we know, we get y minus f of three, f of three, you're gonna take three squared, you get a nine. And then f prime of three we saw was six, x minus three right here. So distribute the six, we get six x minus 18, we're then gonna proceed to add nine to both sides of the equation. We end up with y equals six x minus nine, for which you can see on the graph, exactly that right here, here's our point three comma nine on the graph, illustrated in green is the tangent line. And sure enough, when we take the slope to be six or the y intercept to be negative three, we get the correct tangent line for the graph. Notice how it's just barely touchy, just kissing the function there at the point three comma nine. Let's look at another example. Let's find the tangent line again of y equals x squared when x equals one this time. Well, so we have to look at y minus f of one is equal to f prime at one times x minus one. We just gotta fill in the blanks right here. Well, if you take one squared, you're gonna get a one. If you take two times one, which is what the derivative does here, you're gonna get a two times that by x minus one. Distribute the two, we get two x minus two. And then we have to add one to both sides. Add one, excuse me. And therefore the tangent line is gonna look like y equals two x minus one, which then let's check with our illustration right here. You can see the point of tangency to be one comma one. The y intercept, yeah, that seems like that could be negative one. The y scale is distorted a little bit. This distance right here is five. While this one here is a one. But we do see this exact kissing tangent line right there, which we found the critical part of finding the tangent line, the hard part is gonna become from the slope. And we get the tangent slope by computing the derivative and evaluating the derivative at the specific number.