I’ve told you multiple times that the derivative of a curve at a point is the slope of the tangent line, but our friend [? Akosh ?] sent me a problem where it actually wants you to find the equation of the tangent line. And I realize, I’ve never actually done that.
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Equation of a tangent line Taking derivatives Differential Calculus
So it’s worthwhile. So let’s do that. So it says, find the equation of the tangent line to the function f of x is equal to x e to the x at x is equal to 1.
So let’s just get an intuition of what we’re even looking for. So this function is going to look something like, I actually graphed it, because it’s not a trivial function to graph. So this is x e to the x, this is what it looks like.
I’m just using a graphing calculator, and you can see, I just typed it in. And what this is asking us, is ok. At the point, x is equal to 1. So this is the point x is equal to one.
So f of x is going to be someplace up here, and actually, f of x is going to be equal to e, right? Because f of 1 is equal to what? 1 times e to the 1. So it equals e. So we’re saying at the point, at the point 1 comma e, so at the point 1 comma 2.71, whatever, whatever.
So that’s what point? That’s this point. So it’s right here. 2 point, this is e right here, the point 1 comma e.
So we want to do is figure out the equation of the line tangent to this point. So what we’re going to do, is we’re going to solve it by figuring out its slope, which is just the derivative at that point. So we have to figure out the derivative at exactly this point. And then we use what we learned from algebra 1 to figure out its equation, and we’ll graph it here, just to confirm that we actually figured out the equation of the tangent line.
So the first thing we want to know is the slope of the tangent line, and that’s just the derivative at this point. When x is equal to 1, or at the point 1 comma e. So what’s the derivative of this? So f prime of x.
f prime of x is equal to, well, this looks like a job for the product rule. Because we know how to figure out the derivative of x, we know how to figure out the derivative of e to the x, and they’re just multiplying by each other. So the product rules help us.
The derivative of this thing is going to be equal to the derivative of the first expression of the first function. So the derivative of x is just 1, times the second function, times e to the x, plus the first function, x, times the derivative of the second function. So what’s the derivative of e to the x? And that’s what I find so amazing about the number e, or the function e to the x, is that the derivative of e to the x is e to the x.
The slope at any point of this curve is equal to the value of the function. So this is the derivative. So what is the derivative of this function at the point x is equal to 1, or at the point 1 comma e? So we just evaluate it.
We say f prime of 1 is equal to 1 time e to the 1 plus 1 times e to the 1, well, that’s just equal e plus e. And that’s just equal to 2 e. And you know, we could figure out what that number, e is just a constant number, but we write e because it’s easier to write e than 2.7 et cetera, and an infinite number of digits, so we just write 2e.
So this is the slope of the equation, or this is the slope of the curve when x is equal to one, or at the point 1e, or 1 f of 1. So what is the equation of the tangent line? So let’s go ahead and take this form, the equation’s going to be y is equal to, I’m just writing it in the, you know, not the point slope, the mx plus b form that you learned in algebra. So the slope is going to be 2e.
We just learned that here. That’s the derivative when x is equal to 1. So 2e times x plus the y-intercept. So if we can figure out the y-intercept of this line, we are done.
We have figured out the equation of the tangent line. So how do we do that? Well, if we knew a y or an x where this equation goes through, we could then solve for b. And we know a y and x that satisfies this equation.
The point 1 comma e. The point where we’re trying to find the tangent line, right? So this point, 1 comma e, this is where we want to find the tangent line. And by definition, the tangent line is going to go through that point. So let’s substitute those points back in here, or this point back into this equation, and then solve for b.
So y is equal to e, is equal to 2 e, that’s just the slope at that point, times x, times 1, plus b. It might confuse you, because e, you’ll say, oh, e, is that a variable? No, it’s a number, remember, it’s like pi. It’s a number. You can substitute 2.
7 whatever there, but we’re not doing that, because this is cleaner. And let’s solve. So you get e is equal to 2e plus b. Let’s subtract 2e from both sides.
You get b is equal to e minus 2e. b is equal to minus e. Now we’re done. What’s the equation of the tangent line? It is y is equal to 2 times e x plus b.
But b is minus e, so it’s minus e. So this is the equation of the tangent line. If you don’t like these e’s there, you could replace that with the number 2.7 et cetera, and this would become 5 point something, and this would just be minus 2.
7 something. But this looks neater. And let’s confirm. Let’s use this little graphing calculator to confirm that that really is the equation of the tangent line.
So let me type it in here. So it’s 2, 2 times e times x, right, that’s 2ex minus e. And let us graph this line.
There we go. It graphed it. And notice that that line, that green line, I don’t know if you can, maybe I need to make this bigger for it to show up, bolder. I don’t know if that helps.
But if you look here, so this red, this is our original equation, x e to the x, that’s this curve. We want to know equation of the tangent line at x is equal to 1. So it’s the point x is equal to 1. And when x is equal to 1, f of x is e, right, you can just substitute back into the original equation to get that.
So this is the point, 1 comma e. So the equation of the tangent line, its slope is going to be the derivative at this point. So we solved the derivative of this function, and evaluated it at x is equal to 1. That’s what we did here.
We figured out the derivative, evaluated x equals 1. And so we said, OK, the slope. The slope at when x is equal to 1 and y is equal to e, the slope at that point is equal to 2e. And we figured that out from the derivative.
And then we just used our algebra 1 skills to figure out the equation of that line. And how did we do that? We knew the slope, because that’s just the derivative at that point. And then we just have to solve for the y-intercept.
And the way we did that is we said, well, the point 1 comma e is on this green line as well. So we substituted that in, and solve for our y-intercept, which we got as minus e, and notice that this line intersects the y-axis at minus e, that’s about minus 2. And there we have it.
We have shown that, and visually, it shows that this is the tangent line. Anyway, hope you found that vaguely useful. If you did, you should thank [? Akosh ?] for being unusually persistent, and having me do this problem.
See you in the next video.
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