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I've told you multiple times that the derivative of a curve

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at a point is the slope of the tangent line, but our

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friend [? Akosh ?]

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sent me a problem where it actually wants you to find the

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equation of the tangent line.

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And I realize, I've never actually done that.

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So it's worthwhile.

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So let's do that.

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So it says, find the equation of the tangent line to the

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function f of x is equal to x e to the x at x is equal to 1.

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So let's just get an intuition of what we're even looking for.

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So this function is going to look something like, I actually

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graphed it, because it's not a trivial function to graph.

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So this is x e to the x, this is what it looks like.

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I'm just using a graphing calculator, and you can

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see, I just typed it in.

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And what this is asking us, is ok.

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At the point, x is equal to 1.

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So this is the point x is equal to one.

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So f of x is going to be someplace up here, and

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actually, f of x is going to be equal to e, right?

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Because f of 1 is equal to what?

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1 times e to the 1.

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So it equals e.

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So we're saying at the point, at the point 1 comma e, so at

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the point 1 comma 2.71, whatever, whatever.

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So that's what point?

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That's this point.

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So it's right here.

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2 point, this is e right here, the point 1 comma e.

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So we want to do is figure out the equation of the

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line tangent to this point.

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So what we're going to do, is we're going to solve it by

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figuring out its slope, which is just the derivative

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at that point.

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So we have to figure out the derivative at

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exactly this point.

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And then we use what we learned from algebra 1 to figure out

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its equation, and we'll graph it here, just to confirm that

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we actually figured out the equation of the tangent line.

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So the first thing we want to know is the slope of the

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tangent line, and that's just the derivative at this point.

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When x is equal to 1, or at the point 1 comma e.

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So what's the derivative of this?

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So f prime of x.

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f prime of x is equal to, well, this looks like a

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job for the product rule.

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Because we know how to figure out the derivative of x, we

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know how to figure out the derivative of e to the x, and

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they're just multiplying by each other.

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So the product rules help us.

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The derivative of this thing is going to be equal to the

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derivative of the first expression of the

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first function.

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So the derivative of x is just 1, times the second function,

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times e to the x, plus the first function, x, times the

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derivative of the second function.

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So what's the derivative of e to the x?

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And that's what I find so amazing about the number e, or

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the function e to the x, is that the derivative of e

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to the x is e to the x.

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The slope at any point of this curve is equal to the

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value of the function.

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So this is the derivative.

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So what is the derivative of this function at the point x

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is equal to 1, or at the point 1 comma e?

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So we just evaluate it.

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We say f prime of 1 is equal to 1 time e to the 1 plus 1 times

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e to the 1, well, that's just equal e plus e.

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And that's just equal to 2 e.

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And you know, we could figure out what that number, e is just

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a constant number, but we write e because it's easier to write

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e than 2.7 et cetera, and an infinite number of digits,

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so we just write 2e.

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So this is the slope of the equation, or this is the slope

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of the curve when x is equal to one, or at the point

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1e, or 1 f of 1.

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So what is the equation of the tangent line?

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So let's go ahead and take this form, the equation's going to

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be y is equal to, I'm just writing it in the, you know,

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not the point slope, the mx plus b form that you

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learned in algebra.

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So the slope is going to be 2e.

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We just learned that here.

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That's the derivative when x is equal to 1.

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So 2e times x plus the y-intercept.

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So if we can figure out the y-intercept of this

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line, we are done.

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We have figured out the equation of the tangent line.

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So how do we do that?

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Well, if we knew a y or an x where this equation

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goes through, we could then solve for b.

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And we know a y and x that satisfies this equation.

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The point 1 comma e.

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The point where we're trying to find the tangent line, right?

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So this point, 1 comma e, this is where we want to

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find the tangent line.

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And by definition, the tangent line is going to

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go through that point.

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So let's substitute those points back in here, or this

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point back into this equation, and then solve for b.

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So y is equal to e, is equal to 2 e, that's just the slope at

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that point, times x, times 1, plus b.

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It might confuse you, because e, you'll say, oh, e,

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is that a variable?

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No, it's a number, remember, it's like pi.

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It's a number.

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You can substitute 2.7 whatever there, but we're not doing

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that, because this is cleaner.

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And let's solve.

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So you get e is equal to 2e plus b.

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Let's subtract 2e from both sides.

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You get b is equal to e minus 2e.

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b is equal to minus e.

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Now we're done.

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What's the equation of the tangent line?

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It is y is equal to 2 times e x plus b.

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But b is minus e, so it's minus e.

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So this is the equation of the tangent line.

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If you don't like these e's there, you could replace that

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with the number 2.7 et cetera, and this would become 5 point

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something, and this would just be minus 2.7 something.

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But this looks neater.

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And let's confirm.

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Let's use this little graphing calculator to confirm that that

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really is the equation of the tangent line.

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So let me type it in here.

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So it's 2, 2 times e times x, right, that's 2ex minus e.

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And let us graph this line.

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There we go.

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It graphed it.

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And notice that that line, that green line, I don't know if you

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can, maybe I need to make this bigger for it to

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show up, bolder.

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I don't know if that helps.

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But if you look here, so this red, this is our original

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equation, x e to the x, that's this curve.

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We want to know equation of the tangent line

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at x is equal to 1.

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So it's the point x is equal to 1.

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And when x is equal to 1, f of x is e, right, you can just

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substitute back into the original equation to get that.

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So this is the point, 1 comma e.

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So the equation of the tangent line, its slope is going to be

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the derivative at this point.

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So we solved the derivative of this function, and evaluated

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it at x is equal to 1.

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That's what we did here.

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We figured out the derivative, evaluated x equals 1.

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And so we said, OK, the slope.

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The slope at when x is equal to 1 and y is equal to e, the

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slope at that point is equal to 2e.

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And we figured that out from the derivative.

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And then we just used our algebra 1 skills to figure out

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the equation of that line.

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And how did we do that?

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We knew the slope, because that's just the derivative

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at that point.

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And then we just have to solve for the y-intercept.

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And the way we did that is we said, well, the point 1 comma e

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is on this green line as well.

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So we substituted that in, and solve for our y-intercept,

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which we got as minus e, and notice that this line

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intersects the y-axis at minus e, that's about minus

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2.7 something.

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And there we have it.

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We have shown that, and visually, it shows that

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this is the tangent line.

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Anyway, hope you found that vaguely useful.

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If you did, you should thank [? Akosh ?]

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for being unusually persistent, and having me do this problem.

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See you in the next video.