![]() Have just said the rest of these terms all go to zero.The little trick he uses is not sufficient enough for me to prove to myself the product rule. So if you just knew that youĬould have done this, but it's much better to do it with That are relevant to this proof because all the other terms getĬanceled out when delta x approaches zero. To know a little bit- but if you did a littleĮxperimentation you would realize that whenever youĮxpand a plus b to the nth power, first term is going toīe a to the n, and the second term is going to be plus n a to Knew in any binomial expansion- I mean, you'd have Really didn't even have to know the binomial theorem because we Out is, you know I said that we had to know theīinomial theorem. Works for all real numbers and the exponent. We just proved the derivativeįor any positive integer when x to the power n, where n So n factorial dividedīy n minus 1 factorial, that's just equal to n. 10 factorial divided byĩ factorial that's 10. Or if I have 3 factorialĭivided by 2 factorial, that's just 3, you can work it out. If I have 7 factorial dividedīy 6 factorial, that's just 1. And what's n choose 1? That equals n factorial over 1įactorial divided by n minus 1 factorial times x This is equal to n choose 1 of x the n minus 1. Of the other n minus 1 terms, they're all zeros. We divided by delta x has a delta x in it. This first term has noĭelta x in it, but every other term does. ![]() Pretty much every term that has a delta x in it, Then what are we doing now? Remember, we're taking So the last term becomes nĬhoose n, x to the zero is 1, we can ignore that. And then the last term isĭelta x to the n, but then we're going to divide Of terms, we're going to divide all of them by delta x. We divide by delta x we just get a delta x here. What's delta x divided byĭelta x, that's just 1. So that is equal to the limitĪs delta x approaches zero of, so we divide the top and theīottom by delta x, or we multiply the numerator This is the same thingĪs 1 over delta x times this whole thing. The numerator and the denominator by delta x. If we look at every term here,Įvery term in the numerator has a delta x, so we can divide The n here, and at the very end we subtract out an x to the Green that's x plus delta x to the n, so minus Is going to be 1- it would be n choose n which is 1. They are and, of course, the last digit we just keep adding Through all the digits but the binomial theorem tells us what Of the digits, and in this proof we don't have to go ![]() n choose 1 of x to the n minusġ delta x plus n choose 2 x to the n minus 2, that's x n Theorem if this is looks like latin to you and youĭon't know latin. I'm just going to do the numerator- x to the And what's theīinomial theorem? This is going to be equal to. ![]() This one- the limit as delta x approaches zero. That this is equal to- I'm going to need some space for And if you don't know theīinomial theorem, go to my pre-calculus play listĪnd watch the videos on the binomial theorem. Theorem we can figure out what the expansion of x plusĭelta x is to the nth power. Situation is x plus delta x to the nth power, right? Minus f of x, well f of x How do we take the derivative? Well, what's the classicĭefinition of the derivative? It is the limit as delta xĪpproaches zero of f of x plus delta x, right? So f of x plus delta x in this Let's take the derivativeīinomial theorem, we have the tools to do it. Think now is good time to do the proof of the derivative The binomial theorem, so I think, now that they're done, I Now divide by h and apply some induction to arrive at the desired conclusion. Letting b = x+h and a = x in the formula above, one gets (To prove this identity, simply expand the right hand side, and note that most of the terms will cancel - alternatively, prove it by induction.) There is yet another proof relying on the identity We will prove that it holds for n + 1 as well. Suppose the formula d/dx xⁿ = nxⁿ⁻¹ holds for some n ≥ 1. For instance, using induction and the product rule will do the trick: What's more, one can prove this rule of differentiation without resorting to the binomial theorem. I am sure you can find a proof by induction if you look it up. You don't have to know any combinatorics to understand the binomial formula - it is purely algebraic, it just happens to have a combinatorial interpretation as well. ![]()
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