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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 22: Quotient rule

# Tangent to y=𝑒ˣ/(2+x³)

Sal finds the equation of the line tangent to the curve y=eˣ/(2+x³) at the point (1,e/3). Created by Sal Khan.

## Video transcript

We have the curve y is equal to e to the x over 2 plus x to the third power. And what we want to do is find the equation of the tangent line to this curve at the point x equals 1. And when x is equal to 1, y is going to be equal to e over 3. It's going to be e over 3. So let's try to figure out the equation of the tangent line to this curve at this point. And I encourage you to pause this video and try this on your own first. Well, the slope of the tangent line at this point is the same thing as the derivative at this point. So let's try to find the derivative of this or evaluate the derivative of this function right over here at this point. So to do that, first I'm going to rewrite it. You could use the quotient rule if you like, but I always forget the quotient rule. The product rule is much easier for me to remember. So I can rewrite this as y is equal to-- and I might as well color code it-- is equal to e to the x times 2 plus x to the third to the negative 1 power. And so the derivative of this, so let me write it here. So y prime is going to be equal to the derivative of this part of it, e to the x. So the derivative of e to the x is just e to the x. Just let me write that. So we're going to take the derivative of it. And that's what's amazing about e to the x, is that the derivative of e to the x is just e to the x times this thing. So times 2 plus x to the third to the negative 1. And then to that we're going to add this thing. So not its derivative anymore. We're just going to add e to the x times the derivative of this thing right over here. So we're going to take the derivative. So we can do the chain rule. It's going to be the derivative of 2 plus x to the third to the negative 1 power with respect to 2 plus x to the third times the derivative of 2 plus x to the third with respect to x. So this is going to be equal to negative-- I'll write it this way-- negative 2 plus x to the third to the negative 2 power. And then we're going to multiply that times the derivative of 2 plus x to the third with respect to x. Well, derivative of this with respect to x is just 3x squared. And of course, we could simplify this a little bit if we like. But the whole point of this is to actually find the value of the derivative at this point. So let's evaluate. Let's evaluate y prime when x is equal to 1. Y prime of 1 when x is equal to 1. This thing will simplify to-- let's see, this is going to be e times 2 plus 1 to the negative power. So that's just going to be 1/3, right? 2 plus 1 to the negative 1. So that's 3 to the negative 1. That's 1/3. So that's times 1/3 plus e to the first power. Now let's see, what does this do? This part right over here, this is 2 plus 1 to the negative 2 power. So this-- actually, let me-- I don't want to-- so this part right over here is going to be, let's see, this is going to be-- I don't want to make a careless mistake here-- is 3 to the negative 2 power. So 3 squared is 9. 3 to the negative second power is 1/9. And so it's going to be 1/9. Well, you're going to multiply this negative there. So it's negative 1/9. And then we're going to multiply that times 3 times 1. So it's negative 1/9 times 3. Times 3 right over here. So it's negative 3/9 or negative 1/3. So times negative 1/3. And all I did here is I substituted 1 for x and evaluated it. Now this is interesting. I have essentially-- let me rewrite this. This is equal to e over 3 minus e over 3, which is equal to 0. So the slope of the derivative when x is equal to 1 is 0, or the slope of the tangent line is equal to 0. This simplified to a pretty straightforward situation. If I wanted to write a line in slope intercept form, I could write it like this. y is equal to mx plus b, where m is the slope and b is the y-intercept. Now we know that the slope of the tangent line at this point, it has a slope, is 0. So this is going to be 0. So this whole term is going to be 0. So it's just going to have the form y is equal to b. This is just going to be a horizontal line. So what is a horizontal line that contains this point right over here? Well, it contains the value y is equal to e over 3. So this is a horizontal line. It has the same y value the entire time. So if it has the y value e over 3, then we know the equation of the tangent line to this curve at this point is going to be y is equal to e over 3. Another way you could think about this right over here is, well, let's substitute when x is equal to 1. Well, there's not even an x here. But when x is any value, y is equal to e over 3, you get b is equal to e over 3, or you'd get y is equal to e over 3. So it's just a horizontal line. So let's actually visualize this, just to make sure that this actually makes sense. So let me get my graphing calculator out. And so I'm in graphing mode. If you wanted to know how to get there, you literally can just go to graph, y equals, and I will do-- so e to the x power divided by 2 plus x to the third power. That looks right. And I actually set the range ahead of time to save time. So let me graph this. So let's see. Ooh, it does all sorts of interesting things. All right. Oh, look at that. All right, so now we can trace to get to when x is equal to 1. x equals 1. Right over there, you see y is equal to e over 3, which this is kind of its decimal expansion right over here. And it does look like the slope right over here is 0, that the tangent line is just going to be a horizontal line at that point. So that makes me feel pretty good about our answer.