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Démonstration de la formule de dérivation de x puissance n

Transcription de la vidéo
I just did several videos on the binomial theorem, so I think, now that they're done, I think now is good time to do the proof of the derivative of the general form. Let's take the derivative of x to the n. Now that we know the binomial theorem, we have the tools to do it. How do we take the derivative? Well, what's the classic definition of the derivative? It is the limit as delta x approaches zero of f of x plus delta x, right? So f of x plus delta x in this situation is x plus delta x to the nth power, right? Minus f of x, well f of x here is just x to the n. All of that over delta x. Now that we know the binomial theorem we can figure out what the expansion of x plus delta x is to the nth power. And if you don't know the binomial theorem, go to my pre-calculus play list and watch the videos on the binomial theorem. The binomial theorem tells us that this is equal to-- I'm going to need some space for this one-- the limit as delta x approaches zero. And what's the binomial theorem? This is going to be equal to-- I'm just going to do the numerator-- x to the n plus n choose 1. Once again, review the binomial theorem if this is looks like latin to you and you don't know latin. n choose 1 of x to the n minus 1 delta x plus n choose 2 x to the n minus 2, that's x n minus 2, delta x squared. Then plus, and we have a bunch of the digits, and in this proof we don't have to go through all the digits but the binomial theorem tells us what they are and, of course, the last digit we just keep adding is going to be 1-- it would be n choose n which is 1. Let me just write that down. n choose n. It's going to be x to the zero times delta x to the n. So that's the binomial expansion. Let me switch back to minus, green that's x plus delta x to the n, so minus x to the n power. That's x to the n, I know I squashed it there. All of that over delta x. Let's see if we can simplify. First of all we have an x to the n here, and at the very end we subtract out an x to the n, so these two cancel out. If we look at every term here, every term in the numerator has a delta x, so we can divide the numerator and the denominator by delta x. This is the same thing as 1 over delta x times this whole thing. So that is equal to the limit as delta x approaches zero of, so we divide the top and the bottom by delta x, or we multiply the numerator times 1 over delta x. We get n choose 1 x to the n minus 1. What's delta x divided by delta x, that's just 1. Plus n choose 2, x to the n minus 2. This is delta x squared, but we divide by delta x we just get a delta x here. Delta x. And then we keep having a bunch of terms, we're going to divide all of them by delta x. And then the last term is delta x to the n, but then we're going to divide that by delta x. So the last term becomes n choose n, x to the zero is 1, we can ignore that. delta x to the n divided by delta x. Well that's delta x to the n minus 1. Then what are we doing now? Remember, we're taking the limit as delta x approaches zero. As delta x approaches zero, pretty much every term that has a delta x in it, it becomes zero. When you multiply but zero, you get zero. This first term has no delta x in it, but every other term does. Every other term, even after we divided by delta x has a delta x in it. So that's a zero. Every term is zero, all of the other n minus 1 terms, they're all zeros. All we're left with is that this is equal to n choose 1 of x the n minus 1. And what's n choose 1? That equals n factorial over 1 factorial divided by n minus 1 factorial times x to the n minus 1. 1 factorial is 1. If I have 7 factorial divided by 6 factorial, that's just 7. Or if I have 3 factorial divided by 2 factorial, that's just 3, you can work it out. 10 factorial divided by 9 factorial that's 10. So n factorial divided by n minus 1 factorial, that's just equal to n. So this is equal to n times x to the n minus 1. That's the derivative of x to the n. n times x to the n minus 1. We just proved the derivative for any positive integer when x to the power n, where n is any positive integer. And we see later it actually works for all real numbers and the exponent. I will see you in a future video. Another thing I wanted to point out is, you know I said that we had to know the binomial theorem. But if you think about it, we really didn't even have to know the binomial theorem because we knew in any binomial expansion-- I mean, you'd have to know a little bit-- but if you did a little experimentation you would realize that whenever you expand a plus b to the nth power, first term is going to be a to the n, and the second term is going to be plus n a to the n minus 1 b. And then you are going to keep having a bunch of terms. But these are the only terms that are relevant to this proof because all the other terms get canceled out when delta x approaches zero. So if you just knew that you could have done this, but it's much better to do it with the binomial theorem. Ignore what I just said if it confused you. I'm just saying that we could have just said the rest of these terms all go to zero. Anyway, hopefully you found that fulfilling. I will see you in future videos.