Calculus: A Quick Proof of the Power Rule

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Computational Techniques
The Power Rule
A Quick Proof of the Power Rule Page [1 of 1]
Okay. I thought I would just take a minute or two, right now, to give you at least a rough feeling for why you should believe that last spectacular fact that we just discovered--namely the fact--let me write it down for you, just so you can remember what it is--also put it off to my side here. If you have a function that's just x to some power, then the derivative of that function turns out to be a very simple thing. It turns out to be the following. You just take that number upstairs -- that exponent, bring it down as a coefficient, and multiply it by x raised to a new power. The new power being n minus one.
So, take the derivative of x to the n. You bring the n out in front, and then raise x to the n minus one power. Now, why does that always work? Well, your calculus book, by the way, has--I'm sure--a complete proof of this, which at least in the calculus books I've seen, sometimes it's almost incomprehensible.
I just want to give you a rough idea of why you should believe this. By the way, my explanation might be equally as incomprehensible. So, I'm not trying to say that what I'm about to say is going to make any more sense, maybe, than the book. But at least let me say it.
So, how would we even attempt to make a discovery like this? Well, if you want to find the definition--if you want to find the derivative, and you really want to find it for sure, you have to revert back to the definition--that limit thing. So, let me actually begin to set this problem up as though we were going to actually find this by definition. And let me tell you what the real hard part's going to be. The hard part's going to be, we don't know what number n is. If we knew it were a two, we could do it. In fact, we did a lot. If we knew it were a three, it would take us a little while, but we might be able to do that too. But n could be anything, and that's what makes this fact a little bit tricky to prove, but really neat once we got it, because it's used for any n.
So, here we go. Let's try to write this down. What we have here is that the derivative f prime of x would equal - well, the limit as delta x approaches zero of f of x plus delta x. So, I replace this x by an x plus delta x. So, I have to raise all that to the n. Then I subtract off f of x, which is x to the n, and divide all that by delta x. There's the good old fashion definition of the derivative, but now, I'm plugging in the function x to the n. There, you can see that thing to the n minus x to the n.
Now, we have to evaluate the limit. Well, being optimistic and rosy as I always am, let's let delta x go to zero. If we do that, this term drops out right here, and I'm left with x to the n minus x to the n. Well, that's zero divided by zero. Indeterminate form. No surprise.
Okay. Now, how do we tackle this? Well, when we had a two written here, what we did was we actually FOILed thing out and multiplied this by itself and then saw a lot of cancellation occur. If we had n being the number three, you could actually do the same thing, and it might be a sort of interesting exercise if you want to take it on. It's going to take a little bit of algebra, because you have to multiply this thing by itself three times. There's an awful lot of FOILing and untangling to do it.
If it were four, there's a lot of untangling. If it were five or six or a hundred, like in the previous example we looked at, that takes an awful lot of untangling. The point is that it can be untangled. And let me just sort of show you what the untangling would look like, at least the highlights of the untangling. So, I'm actually not going to do it out. There's actually a mathematical theorem called the binomial theorem that lets you untangle it completely, but who cares about that for now?
The point is I want you to visualize in your mind's eye. I could write this down here. I could even have it appearing there, but I don't want to. I want you to actually to think about this in your mind's eye for a second. Imagine having this quantity written out n times. So see the x plus delta x times x plus delta x times x plus delta x, and it's all there written out n times. So, if n were a thousand, you'd see it a thousand times.
Now, to actually FOIL all that out, what you have to do is take every single possible combination of a term here, with a term here, with a term here--all thousand of them multiplied together and added to every other possible combination of taking them. Notice when you view that in your mind's eye, there's only one way of getting a term that just has x in it, with no delta x's in it, and that's if you pick out the first term from the first piece--the x--then the x from the second piece, then the x from the third piece, then the x from the fourth piece and so on. If you do all that, you'll have x. And how many x's will you have? You'll have as many x's as you have terms, and there are n terms here clustered. If you pick the x out of each one of those, you would have n of them.
So, you'd have x to the n, and that's the only way to get a term that has no delta x's in it. Because if you pick out any other way, you'd have to pick out at least one of the delta x's, and you'd have a delta x factor. So, we write that out as x to the n, and then there's other stuff. Now, what's the other stuff?
Well now, this gets actually a little tricky to see, so let me just talk through this and ask myself, "Well, how many times will I just get delta x--one delta x term and then no other delta x terms?" Well, how would you do that? Now, I want you to visualize these things clustered. Well, let's say I pick out that first delta x here in that first little piece. Well, for me to make sure I get no other delta x's, what would I have to pick out of the rest? I would have to pick out of the rest all x's -- x,x,x,x,x,x,x,x. If I picked out another delta x, I'd have two delta x's. I don't want that. I only want one delta x right here.
So, how many x's would there be left over? I have n all together. That first one is a delta x, so the rest, which would be n minus one of them, would be all x's. And so, I would see an x to the n minus one times that first delta x. But of course, there's another way of actually doing the same thing. I could actually have picked that second term to be the delta x piece, and then pick the x's all out of the other ones. That's another combination. I would still have n minus one x's, and I would have one delta x. So, then I would have two of them. So, then I get a third one by picking x, x then a delta x and then the rest x,x,x,x,x.
I keep doing this, and it turns out that there are n of possible ways of doing this. So, in fact, there are n of those things, which I have to add on. That's the next term in untangling this, and the rest of the terms--and they go on for a long time--will always have at least a delta x squared in them. So, I'm going to write in here plus delta x squared times lots of stuff, lots o' stuff. So, that's all the rest of the stuff, which I don't know what it is, but it's very complicated. I had to subtract off this x to the n, and it's all divided by delta x.
Now, this part might be a little bit tricky to sort of follow, and, in fact, I think that it's really hard to follow. Not it might not be; it actually is. I know it is. If you untangle that, it's really hard to think about. So, I want you to think about just picking off the x's first everywhere, and that gives me an x to the n. And then just picking off one delta x and all the rest being x's, and then I get this many of them. And the rest of the stuff looks like this. Maybe that was--that is sort of complicated. Maybe I'm not any better than the book. But, the truth is this is what we get with a lot of complicated stuff here. Let's ignore it.
Notice the happy fact that x to the n minus x to the n cancels just as they always seem to cancel. And what's left--I can factor out a delta x. And when I factor out a delta x, what do I see? Well, if I factor out the delta x, then I'm left with the limit as delta x approaches zero of--and I factor out that delta x. Well, here I'm left with n--x to the n minus one. And then here plus--make a long plus sign there--plus--well, I took out one delta x, and so, I still have a delta x left over, but it's times all this complicated stuff--lot's o' stuff--all over delta x.
So, that delta x is a common factor of the top and the bottom. I cancel it away, and I can take the limit, and if I take the limit, what do you notice? All that lots of stuff gets multiplied by a zero. So, all that stuff goes away in the limit, and all I'm left with is this, and that's the derivative. And that's why this formula holds.
So, I admit that, maybe, this didn't make an awful lot of sense to you. But I wanted to give you at least, in spirit, the opportunity to see that this is not coming out of thin air, and even though we notice that those things are happening in those examples, it always happens. And this is a little, at least a little window into why you could believe it, and I hope you believe it.
Now, what we're going to do is run with it and see a lot of examples. Okay. See you in a bit.