Subtitles section Play video Print subtitles So let's think about how we could find the slope of the tangent line to this curve right over here, so what I have drawn in red, at the point x equals a. And we've already seen this with the definition of the derivative. We could try to find a general function that gives us the slope of the tangent line at any point. So let's say we have some arbitrary point. Let me define some arbitrary point x right over here. Then this would be the point x comma f of x. And then we could take some x plus h. So let's say that this right over here is the point x plus h. And so this point would be x plus h, f of x plus h. We can find the slope of the secant line that goes between these two points. So that would be your change in your vertical, which would be f of x plus h minus f of x, over the change in the horizontal, which would be x plus h minus x. And these two x's cancel. So this would be the slope of this secant line. And then if we want to find the slope of the tangent line at x, we would just take the limit of this expression as h approaches 0. As h approaches 0, this point moves towards x. And that slope of the secant line between these two is going to approximate the slope of the tangent line at x. And so this right over here, this we would say is equal to f prime of x. This is still a function of x. You give me an arbitrary x where the derivative is defined. I'm going to plug it into this, whatever this ends up being. It might be some nice, clean algebraic expression. Then I'm going to give you a number. So for example, if you wanted to find-- you could calculate this somehow. Or you could even leave it in this form. And then if you wanted f prime of a, you would just substitute a into your function definition. And you would say, well, that's going to be the limit as h approaches 0 of-- every place you see an x, replace it with an a. f of-- I'll stay in this color for now-- blank plus h minus f of blank, all of that over h. And I left those blanks so I could write the a in red. Notice, every place where I had an x before, it's now an a. So this is the derivative evaluated at a. So this is one way to find the slope of the tangent line when x equals a. Another way-- and this is often used as the alternate form of the derivative-- would be to do it directly. So this is the point a comma f of a. Let's just take another arbitrary point someplace. So let's say this is the value x. This point right over here on the function would be x comma f of x. And so what's the slope of the secant line between these two points? Well, it would be change in the vertical, which would be f of x minus f of a, over change in the horizontal, over x minus a. Actually, let me do that in that purple color. Over x minus a. Now, how could we get a better and better approximation for the slope of the tangent line here? Well, we could take the limit as x approaches a. As x gets closer and closer and closer to a, the secant line slope is going to better and better and better approximate the slope of the tangent line, this tangent line that I have in red here. So we would want to take the limit as x approaches a here. Either way, we're doing the exact same thing. We have an expression for the slope of a secant line. And then we're bringing those x values of those points closer and closer together. So the slopes of those secant lines better and better and better approximate that slope of the tangent line. And at the limit, it does become the slope of the tangent line. That is the definition of the derivative. So this is the more standard definition of a derivative. It would give you your derivative as a function of x. And then you can then input your particular value of x. Or you could use the alternate form of the derivative. If you know that, hey, look, I'm just looking to find the derivative exactly at a. I don't need a general function of f. Then you could do this. But they're doing the same thing.

B1 US slope tangent derivative line point arbitrary Formal and alternate form of the derivative | Differential Calculus | Khan Academy 2 1 yukang920108 posted on 2022/07/12 More Share Save Report Video vocabulary