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# In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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Okay, So our goal here is to find the equation of the tangent line of f of X passing through the point X equals a So here f of X is one of her ex square. And we want the equation. The tangent line passing through the point X equals one. So what do we need? So we're looking for the equation of a line. We need the slope in the slope of the line. It's just f prime of what? Okay, so the slope of the tangent line is just the derivative and X equals one, so we need to figure out what that is, first of all, and then we need the point on the line or any point on the line to write down the equation of the line. But we know what that ISS. We're just going to use the 0.1 comment ethic. One. We're going to use the point of tange INSEE. So we're going to use the the point where the tangent line just barely touches. Um, the graph of the function and this is actually just one what right, Because about deployment one, I get one. So the real question is what is this. Look, what is the derivative? That's the interesting question here. So let's just write down what the terrific Davis or what the slope of the tangent line is. So it's still limit. His H goes to zero of well, one over one plus h squared minus one over one, squared all over H. Okay, so just in case you're a little bit confused where this came from, remember that the derivative F Prime Obey is the limit is H goes to zero of f of A plus H minus f of a all over H. And so all I did here is just evaluate. So a is one. I just evaluated f at one plus h and then minus f evaluated at one and then divided by h. So this limit as well most of the time be the case when finding a derivative via the limit, it's going to be an indeterminant form. If I just try to plug in H equals zero, I'm going to get 0/0. So I'm always going to have to do more work, and normally that's going to involve just, um, typical algebraic tricks like we've seen before. So what I can do here is well, get a common denominator. So this is one plus age squared. This is just one. So I can write. This is one minus one plus h squared over one plus h squared all over h. So all I did this is just one I just wrote. This is one plus a squared, divided by one plus eight square. And then because I have a common denominator, I could just do the subtraction in the numerator. Okay, So what do we have here? Well, this one plus age squared I can actually expand that out is one plus to H plus H Square. And now I'm taking one and subtracting all of these values. So let's just do that. So one minus one is zero, and then one and then Okay, so one minus 10 and then we're subjecting to age. So we're left with minus two age and then minus h squared. So this is a common mistake. So the temptation is to say, Okay, one minus 10 and then just to h plus age squared. But you got to remember that you're subtracting this entire quantity. So really, the negative is distributed to all of those terms and then divided by one plus h squared all over h. Okay. And I'll skip a little bit here. So the numerator of the numerator has a factor of age, as does the denominator. So I can actually cancel one of those factors of age. So I'm really just left with the limit. His H goes to zero of minus two, minus h over one plus h squared. Okay. And now I could just evaluate. I just plug in h equals zero, Take the limit. So this is going to zero that's gone. And then that h is gone. One squared is just one, so it looks like the slope is just negative to Okay, so the slope is negative two. Then I just use my points look for So it looks like I have Why equals negative, too. Times X minus one plus one. Or I think that ends up being negative. Two x plus three. We'll just leave it like that. But now, because that is the equation of the tangent line and you see, the most important part is finding the derivative. Okay, So negative to we'll see a picture in just a second is like the slope of that function at X equals one. And of course, I'll just writing negative at two X plus three if you prefer soap intercept form. So let's look at a picture real quick. All right, so here is the fruit of our labor. We started with this function one over X Square, and then we were looking specifically at X equals one. Here's our point of tangent C 11 And so we know the tangent line is just gonna barely pass through that point. But we needed to find the slip, right. So then we determined that f prime of one was negative two. So that gave us the slope of the tangent line. Then we just use point slope form to find the equation of the danger line. Negative two x plus three. And the cool thing to do is to use some sort of computer algebra system into plot the line and be amazed when Yeah, actually, this line is tangent to the function. It's just barely touching the function. And at one point it's giving you You see that if I look at the blue function, the slope of that function is deep is well, it's changing, so it's starting really steep. And then it's kind of leveling off. And at X equals one. The slope of the function is equal to the slope of the tangent line, which is negative. Two. So it's telling us how the function is changing at the point X equals one. Okay, so let's do it again. We have so much fun last time. So what do we need? We need the slope, which is going to be f prime of zero. So we'll figure out what that is. That's gonna be most of the work. And then we need a point on the line to get the equation of the tangent line. And we're just going to use that point of Tange Enciso the point where the tangent line touches the graph and that's gonna be zero eth zero, which is going to be zero f of zero is just one. If I plug in zero for X, I get one. So there we go. So let's find the slope. Let's find the derivative So f prime of zero is just going to be square root of H plus one minus one over age, okay? And so why is that? Well, I plug in a plus H for X, but a zero. So I plug in zero plus h. So this is f of zero plus h. Just write that up here. This is f of zero plus age, and then this is f of zero. But we already found the f of zero is one. Okay, so then I just subtract those two quantities and divide by age and again, if I try to plug in H equals zero, I'm going to get an indeterminant form. So I need to do more work. And if you recall with limits, we have a red flag here we have a square root. Let's multiply by the kanji And let me actually not forget the most important part here. That really I need to take the limit is h goes to zero. That's important. Okay, so if I try to take this limit like I said, I'm going to get an indeterminate for him. So I'm gonna multiply by the kanji it so I'm gonna multiply by squared of H plus one plus one on the top and bottom and in the top. That's going to give me a difference of squares. So that's gonna just gonna be H plus one minus one all over H times squared H plus one plus one. And so again, this is just squared of H plus one square, which is H plus one minus one one square, which is mine. Alrighty. And so now notice that the numerator. So this is one minus one. So I'm just going to be left with H. But then that means thes h is will cancel. And, of course, again, in the limit is h goes to zero. I can ignore h equals zero. And so what I'm left with, it's the limit is h goes to zero of one over squared of H plus one plus one. But now I can actually take the limit. So is h goes to zero. This becomes one, so 1/1 plus one is one over to. So the slope of the tangent line at X equals zero is one half already. But we have the y intercept. So actually, the equation the tangent line, it's easy to write down. It's just one half X plus one because the Y intercept is one when x zero, Why is one So this is the equation of the danger line. And again, this is, you know, a good bit of work to take the limit to find the derivative. But the cool thing about this is you get to see it. You get to actually see what you're finding. You're not just doing math Bob Loblaw doing math just to do it. You're doing math and then you get a plot it and you get to see Oh, my goodness, I'm actually finding something geometrically that's really cool. And you're going to see it has so many applications. So let's look at a picture. All right, so here it is. In blue, we have the function square root of X Plus one and we were trying to find the tangent line at the point X equals zero. So that was 01 on the graph of our function, and we found that f prime of zero was one half. So the slope of the function in blue at X equals zero. It's leveling off. So it's starting really steep the slope, and then it's kind of leveling off, and right here it's one half so it's really cool to find these slopes. Find these tangent lines. Look at what's going on. We're really developing these tools to study how functions are changing.

Georgia Southern University

#### Topics

Differentiation

Applications of the Derivative

Integrals

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor