Calculus: Derivative of the Reciprocal Function

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An Introduction to Derivatives
Some Special Derivatives
The Derivative of the Reciprocal Function Page [1 of 4]
This time, I want to take a look at the derivative of a slightly more interesting, slightly more exotic function. Here’s the question, as it would appear, maybe, in a textbook or on a quiz. “Find the equation of the tangent line at x equals three if you’re given that f of x equals one over x. Okay. So, now we’re trying to find the equation of the tangent line. How do you find the equation of a tangent line?
Now, remember here, in our spirit of actually answering the question that’s being asked we have to remember that it’s an equation that we want, and it’s the equation of a tangent line. Okay. So, to find the equation of a line, we need two pieces of information. We need what? Well, we need the slope of the line, and we need a point on the line. We need slope and a point.
Okay. Well, actually, let’s try to draw a picture of this to see, at least, what this thing looks like. You can graph this thing pretty easily over here. You may recall that the graph of one over x is a hyperbola, and you may not. That’s okay, but I remember it’s actually a hyperbola. Looks sort of like this. Let me do it in a different color. There’s the graph of one over x and we’re looking at the point x equals three. So one, two, three. There’s three. And if I go up—you see—put a dot there. I want to look at the tangent line—the tangent line at that point, which would look something like this. So, just draw the line that just grazes it.
Hard to draw lines, by the way, on the fly. I do have a straight edge here, in case you’re wondering, but I want to show you I can also do it on the fly pretty darn straight—pretty darn straight. So, that’s the equation of the line we want. So, we want the equation of this line.
Now, you know what? Just to put a little spice and zap into this problem—this is sort of a boring problem—let me try to make it a little more interesting. Let’s suppose that, in fact, instead of giving us just this function, like, you know one over x. That’s sort of boring. So, forget that. Let’s suppose that, in fact, we were given a gummy worm—a gummy worm that was sort of slithering through space, or slithering through somewhere. And it turns that, during its slither, it actually slithered along this path. And I happen to have a gummy worm here just to illustrate this thing. So, I put the gummy worm right down there.
You know, people think that—you know—these things may be unapplied. But, of course, they are extremely applied, especially if you’re looking at a gummy worm. Look at that. Can you see that? There’s the gummy worm. And now, suppose that you don’t want a gummy worm sitting on your—that’s, by the way, a very long gummy worm. Have you ever seen a gummy worm that long? I think not. But over the information super-highway, folks, anything can happen.
Okay, so you have this gummy worm, but you want it shooed off your desk. So, what are you going to shoo it off with? Well, you’re not going to somehow use your hands, because that’s sort of gross. So you reach for the nearest thing, which, in this case here, is a pepperoni stick. So, you just take out the pepperoni stick, you see. There you go. And well, what you want to do, you want to sort of shoo it off. And what you want to do is you want to shoo it off, but you’re going to hit right there at x equals three.
So, the pepperoni stick, you see, is acting as the tangent line, and it’s just touching the gummy bear just at that point. So, the question is “What is the equation of the pepperoni stick? What is the location of the pepperoni stick so it just touches the gummy bear at the point here, at three?” That’s the problem. _______.
That’s one tasty tangent line. Okay. Well, so how do we do this? Well, let’s try it. What we first want to do is – well, you can either find the slope of this line—the slope of the pepperoni stick or we could find a point on the pepperoni stick.
Let’s think about the second question first, because that’s actually, I think, a little easier. Tell me a point that actually is on the pepperoni stick. Do you know any points on the pepperoni stick? Well, I don’t know any of the points over here or over here, but look, there’s a dot right there. And it turns out that that point, the point of tangency, is a point that’s shared by both the pepperoni and the gummy worm. Because since they’re tangent—those two things are tangent at that point—that means, in particular, they must touch at that point.
So that’s a point in common. And what that means is that the point—whatever the point is on the gummy worm—whatever that height is, at x equals three, of the gummy worm—it has to be the identical height of the pepperoni stick. So, what is the height of the gummy worm there? That’s actually pretty easy to figure out. I just take three and plug it into the function. When I put three into here, I see the number one third. So, this turns out to be one third. So, therefore, three comma one third must be on the pepperoni stick. And so, three comma one third is a point on the pepperoni stick.
Now, what we need—need now is the slope of the pepperoni stick. Well, how do we find the slope of this line? Well, we don’t know a lot about the pepperoni stick, except we do know that the pepperoni stick is tangent to the gummy bear. So now, what we have to do is find the slope of a tangent pepperoni stick. Well, how do you find slopes of tangents? Well, we know how to do that, because I remind you, once again, that—well—that they’re right here.
The derivative of a function represents the slope of the tangent line. This is a typo here. This should say tangent pepperoni stick. But that’s okay, I think you understand that. So, what I need to do is, actually, find the derivative of this function, and then, evaluate that derivative. The derivative, remember, is a machine that spits out slopes, and that’s what I want here. I would take that derivative, and I would, then, plug in x equals three and see what the slope of the pepperoni stick is at that point.
So, we find this, need the derivative, need f prime of x. And this is exactly how these problems always go. We need to find the point on the tangent pepperoni stick, which is always going to be the point of tangency, and then we need to find the slope, which requires us to take the derivative.
Okay. Well now, that’s all just a set up. Now, I’m going to actually look at this question. We need to find the derivative. So, this is now going to be a derivative problem. We have to go off, and now find the derivative. Look how many steps are involved in finding the equation for this tangent pepperoni.
So, let’s now—I’ll move the picture. If you want to look at the picture for a little while longer, I’ll put the picture over there for a second. You can take a look at that. But for now, what we want to do is we’re given this function f of x. I’m going to remind you, it’s the function one over x. That’s the gummy worm function. And I want to find the derivative. Why do I want to find the derivative? I want to find the derivative, because that gives me slopes of tangent lines, or pepperoni sticks.
Okay. Well, what’s the recipe? Well, I remind you the recipe is right here. I hope you’re getting a sense of this, by the way, because this is really an important formula. One that I really hope you understand and grow to like. I know it looks sort of scary and long, but the more you work with it, the more I think you’ll make it your own.
So now, let’s take the derivative, and plug in to this function. So, I’ll see f—do this over here so you can watch it a little bit more if you want—f prime of x equals the limit as delta x approaches zero of—well, f of x plus delta x. Now, what is f of x plus delta x? Well, f of x is one over x. So what this means is that whatever I put in here, if you want to see what x does to it, you just take the reciprocal of the whole thing. So, I’ve got to take the reciprocal of x plus delta x. So, I write here one over x plus delta x—that value there is equal to this—minus f of x. Well, that’s just one over x. But don’t forget that’s all divided by that delta x.
If I bring this over here, you can really see there’s a similarity to here. All I did here was insert all the information knowing the function. This is the general definition of a derivative, but for this particular function, you have to plug in, and you get this.
Okay. Well now, we’re off and rolling. First thing I would always check, of course, is let delta x go to zero and see what we get. Of course, we all know the answer. The answer’s going to be an indeterminate form. But, I always check this, because maybe something nice would happen one day, and there won’t be an indeterminate form.
Okay, if delta x goes to zero, that term drops out right there, and then I see one over x minus one over x. That’s zero on top and then zero on the bottom—indeterminate form. No surprise. What to do? Well now, you might want to remember back to those cool tricks we talked about earlier on when we were talking about limits. I see a very complicated looking fraction. What makes this so complicated? Well, it’s fractions of top of fractions.
Let’s try to make it into one fraction. And we saw this technique earlier, and I want to now use it here. I’m just going to get a common denominator and combine these fractions. Now, a great guess for the common denominator might be x plus delta x and that sounds like an okay guess. Turns out, though, it’s not a really great guess, because it’s difficult for me to find a multiple to multiply top and bottom here and something to multiply top and bottom here to make them both x plus delta x. I just can’t add a delta x. I can multiply top and bottom.
So, actually, the common denominator will be x times x plus delta x. So, I have to multiply top and bottom here by x, and I have to multiply top and bottom here by x plus delta x, and then all that is over delta x. Looks like a teeter-totter doesn’t it. It sort of looks like it’s going to go either way, but it’s just holding on it’s balanced perfectly. These now have the same bottom, so I’m allowed to subtract the top, keep the bottom, and if I do that, what do I see? I see the limit as delta x approaches zero. And if I subtract the tops, I see an x minus x plus delta x, and it’s all divided by that bottom, x times x plus delta x. But then, don’t forget it’s all divided by delta x. So I combined that to get this divided by that.
Are you happy? No. I hope you’re not happy, because I made one of those classic mistakes. The mistake was I have to subtract everything, so I have to remember to put the parentheses here. In particular, at some point, I am going to have to distribute that negative sign throughout. That’s going to be important.
Okay. Well, let’s now take a look and see what this equals. This equals the limit as delta x approaches zero. Well, notice that there’s a cancellation – x minus x actually drops out and I’m just left with a negative delta x on the top. So, negative delta x on the top, all divided by x times x plus delta x, and that’s all divided by delta x.
What do you do? Well, now this is a fraction. I can visualize this over one, and I invert and multiply. So, I see now one over delta x, and so I see minus delta x over x times x plus delta x—a lot of stuff there. Notice I don’t distribute down there. You might remember what I said earlier that I never distribute denominators. I always keep them factored. Only distribute the numerators. So, in other words, I just keep carrying that along, and I have to multiply that by the reciprocal of this. Watch it—whoop. So it’s one over delta x. And now, happily, I finally see the zero over zero, and I can cancel them away.
Now, what am I left with? I’m left with a negative—don’t forget that negative sign there. Don’t cancel that. Negative one—this equals the limit as delta x goes to zero of negative one divided by—and I’m left with x times x plus delta x. Now, I’ve simplified that quite a bit. I can take the limit of this. So, what happens if delta x goes to zero? Well, then this term, right here—that delta x term—goes to zero. And what remains? Well, what remains is the negative one on top and the one on the bottom. I see an x times x plus zero. Well that’s just x times x, which is x squared.
So, what I’ve just discovered is that the derivative f prime of x—the derivative of this function way back here of one over x—turns out to be negative one over x squared. That’s what I just figured out.
A lot of work. Notice that there’s a little bit of dejá vous if you saw our early discussion about finding limits of things that look like this. It’s actually—the example we did back then was one that was almost exactly like this problem. So again, once you set the right limit, if you’re careful with the limits, it all falls out, and we get our answer.
Okay. What does this answer represent? Well, let me remind you, that the answer represents a slope. Right? The derivative represents the slope of the tangent line, and that was the missing ingredient in our question—our question of finding the equation of the tangent line of the gummy bear. So, let me now move this over to here. Let me just record that last fact that we just discovered that’s the derivative—the little machine that produces slopes—is equal to that.
Okay. Well now, where are we? Well, we want to now find the slope of the tangent line. And in fact, let me just quickly recapture that picture for you. I can draw it, in fact, over there. It’s a cleaner picture there, but with the gummy bear, just so you can see what’s going on here. Let me remind you what’s going on. We’ve got that curve that looks like this. Here’s three. We already saw this point is one third, because we plugged in to the one over x gummy bear function. And then, the pepperoni stick is right here like that. I guess, actually, I really can’t put it right at the same point, because it has to be sort of next to that. That’s how it really would look. See how it looks just like that. Just at the same point, I could sort of put it like that, but you can pretend.
Okay. Now, how do I find the slope of the tangent line there? I just plug in three into this formula, because this is the machine that produces slopes. That’s minus one over three squared, which is nine—so minus one ninth. Is that at least a sensible answer – in particular, that negative sign? If a slope is negative, what does that mean? How is it pitched? It’s pitched downward. And notice this, in fact, is a downward-pitched pepperoni.
So, in fact, this at least makes sense. If we would have got an answer of plus a ninth, that wouldn’t make sense, because a slope of plus a ninth would be something like that. Well, that’s not tangent. So, in fact, we could have seen, just by thinking about it, the answer should be negative. And, in fact, notice that this quantity here is always negative. Because if you put any number in here, whether it’s positive or negative, after I square it, it becomes positive.
So, this negative sign makes this negative. And notice that no matter where I put the pepperoni, the pepperoni will be negatively sloped. So, in fact, that’s sort of interesting that we now see that the pepperoni’s always negatively sloped, and that follows from the fact that the slope of all the tangents are going to be negative. That’s sort of neat little extra fact we got for free.
But, whether it’s neat or extra or free, the point is we have to answer the question that someone asked us, and that was to find the equation of the tangent line. We now have the slope. We already found a point on it, and you can use the y-intercept form. But if you’re like me, Mr. Lazy—I’m sorry, Professor Lazy—I’m just going to use the point/slope form.
I’ve got a point. I’ve got a slope. And so, I don’t have to think any more about this. I can just very nicely say that the formula is y minus one third—that’s the y-value here—equals minus one ninth times x minus the known x-value, which is three. And that’s the answer. Look how nicely using this equation—form for the equation of a line, it just pops right out.
Great. Great. Well, a job well done. And here’s looking at you. See you soon.