We dive into proving the formula for the derivative of x^n by skillfully applying the binomial theorem. Together, we expand (x + Δx)^n, simplify the expression, and take the limit as Δx approaches zero to reveal the power rule for derivatives. Created by Sal Khan.
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- In the second step, I thought the definition of a derivative was (f(x+∆x)-f(x))/∆x. But in this video it shows ((x+∆x)^n-x^n)/∆x. Why is this?(11 votes)
- Using f(x) notation generalized the definition of the derivative. Now, we are plugging in what are actual f(x) is. In this case, f(x)=x^n so f(x+∆x)=(x+∆x)^n and so on.(22 votes)
- What does the limit (delta)x approaches zero mean? Why does Sal say that (delta)x approaches zero?(5 votes)
- Delta X is the change in X, to take the average slope of a function (over the range of delta x) you divide delta Y by Delta X.
To take the slope of a function at any point (i.e. the derivative) you take the limit as delta X approaches 0.(7 votes)
- Why is the derivative of x^n = nx^n-1, I mean I know Sal proved it above but I was wondering since the answer is so neat is there a physical or intuitive reason as to why it is so neat?(5 votes)
- Because you are looking at a tanget line of a very well defined curve. That makes it nice and neat.(6 votes)
- When Sal is cancelling out all the terms while he is taking the limit, I noticed that if n=1 then that last term would be ∆x^0. But since he was turning all the ∆x's into zero, if n=1 then the last term would be 0^0, which is undefined, so...how does that work out?(5 votes)
- *That's a good question.*
Well, ∆x is tending to zero, which means it is infinitely small, but not 'zero'. It might seem confusing, but it's sort of like predicting what would happen if ∆x became zero (by taking ∆x to be something that's really really really ... (infinite times) ... really small).(2 votes)
- Where can I find Leibniz Rule in KHANACADEMY.ORG's library?(3 votes)
- According to Wikipedia, there's two Leibniz rules: http://en.wikipedia.org/wiki/Leibniz%27s_rule. The first one is the product rule and is covered here https://www.khanacademy.org/math/calculus/differential-calculus/product_rule/v/product-rule (not HD) and here https://www.khanacademy.org/math/calculus/differential-calculus/product_rule/v/applying-the-product-rule-for-derivatives (HD).
I don't think he has anything for the other Leibniz rule (i.e. the Leibniz integral rule) yet.(2 votes)
- How would you differentiate x^x w.r.t. x?
In case of x^x, both the base and the index are variables
So do we use dx^n/dx or do we use d(a^x)/dx?(2 votes)
- For those of us who skipped the Binomial Theorem in the Algebra II playlist because they don't know any combinatorics, how important is this proof? Can it be safely skipped for the time being? Building up one's knowledge of combinatorics to the point where one can understand and apply the Binomial Theorem feels like a lot of work...(2 votes)
- 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. I am sure you can find a proof by induction if you look it up.
What's more, one can prove this rule of differentiation without resorting to the binomial theorem. For instance, using induction and the product rule will do the trick:
Base case n = 1
= lim (h → 0) [(x + h) - x]/h
= lim (h → 0) h/h
d/dx x¹ = 1x⁰.
Suppose the formula
d/dx xⁿ = nxⁿ⁻¹holds for some
n ≥ 1. We will prove that it holds for
n + 1as well. We have
xⁿ⁺¹ = xⁿ · x. By the product rule, we get
= d/dx (xⁿ · x)
= [d/dx xⁿ]·x + xⁿ·[d/dx x]
= nxⁿ⁻¹ · x + xⁿ · 1
= nxⁿ + xⁿ
= (n + 1)xⁿ.
This completes the proof.
There is yet another proof relying on the identity
(bⁿ - aⁿ)
= (b - a)[bⁿ⁻¹ + bⁿ⁻²a + bⁿ⁻³a² + … + b²aⁿ⁻³ + baⁿ⁻² + aⁿ⁻¹].
(To prove this identity, simply expand the right hand side, and note that most of the terms will cancel - alternatively, prove it by induction.)
b = x+hand
a = xin the formula above, one gets
(x+h)ⁿ - xⁿ
= h[(x+h)ⁿ⁻¹ + x(x+h)ⁿ⁻² + … + xⁿ⁻²(x+h) + xⁿ⁻¹].
Now divide by
hand apply some induction to arrive at the desired conclusion.(3 votes)
- isn't it simpler to use power rule(0 votes)
- This is supposed to be a proof of the power rule. Using the power rule to prove the power rule would be circular reasoning. Also, one can not logically use the power rule without first proving it.(6 votes)
- Can someone please give the link of generalization of (x+dx)^n - x^n/dx from derivative formula?(2 votes)
- There is no generalization for what you just said. However, when dx->0, the whole expression simplifies to nx^n-1.(1 vote)
- How would you prove that the derivative is nx^(n-1) if n is complex?(1 vote)
- Assuming x is a real number, you can write x^n = e^(log x^n) = e^(n log x), and differentiate to get
n/x e^(n log x) = n/x x^n = nx^(n-1). n is just a constant, so it doesn't matter whether it's real or complex.(2 votes)
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 1. 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.