For positive integer n, you can prove it by induction and with the help of the product rule. For negative integers, apply the quotient rule and the result for positive integers. For general n as a real number or even a complex number, it is perhaps more elegant to define exp and log first, develope their derivative formulas, and then define power x^n = exp(n log x) and apply the chain rule.
Nice video. Please note that at 5:03, after the change of variable from k (which goes from 0 to n) to j=k+1, the new sum should be from 1 to n+1 (not n). You then separate off the term for j=n+1, which is hⁿ⁺¹, and then continue as before (but adding +hⁿ⁺¹ at the end of each line). At 8:55, when you put everything together, you now do indeed get the sum for k=0 to n+1 as stated.
@dibeos No problem. I appreciate the effort you guys put into the video (and making the PDF available, as usual). You might be interested in seeing this related proof by Dr Barker: "Differentiating x^{1/n} from First Principles without Power Series" ruclips.net/video/-A7T8_5zhoc/видео.html
I was literally just thinking about this last night right before i slept... crazy coincidence that you guys uploaded a video on this exact topic the next day
Wonderful video 🙌 By far the trickiest part of this process is proving the binomial theorem with induction!! my favorite proof of the power rule (or at least the one I have memorized) is MUCH shorter: y = xⁿ ln(y) = nln(x) If we implicitly differentiate both sides we get: y'/y = n(1/x) Now just solve for y' and sub in y = xⁿ y' = (n/x)y = (n/x)xⁿ = nx^(n-1) Ok, I know using implicit differentiation and the derivative of the log is a little heavy handed but it leads to such a simple proof 😅 Great video!
@@graf_paper thanks! yeah, this proof is much shorter, but you do need to rely on the fact that one knows how to differentiate ln(X). I guess that’s the trade-off. Anyway, thanks for sharing. I didn’t really know hahah and yes, you’re right, I could title the video: how to prove the binomial theorem since most of the video is about that 😎
I think the approach taken here could potentially give some people, especially those new to these concepts, the impression that math/derivatives/proofs are always much more complicated than they need to be. Really, it's not too hard to show using just the distributivity of multiplication that (x+h)^n = x^n + hnx^(n-1) + other terms with powers of h of degree at least 2. The rest of the proof follows easily. I'm sure you're already aware of this. I'm just against the idea of overcomplicating simple stuff that a student might come across early into his learning journey. Concepts such as the binomial theorem can be introduced later in other contexts where they're more relevant.
For positive integer n, you can prove it by induction and with the help of the product rule.
For negative integers, apply the quotient rule and the result for positive integers.
For general n as a real number or even a complex number, it is perhaps more elegant to define exp and log first, develope their derivative formulas, and then define power x^n = exp(n log x) and apply the chain rule.
Nice video. Please note that at 5:03, after the change of variable from k (which goes from 0 to n) to j=k+1, the new sum should be from 1 to n+1 (not n). You then separate off the term for j=n+1, which is hⁿ⁺¹, and then continue as before (but adding +hⁿ⁺¹ at the end of each line). At 8:55, when you put everything together, you now do indeed get the sum for k=0 to n+1 as stated.
@@MichaelRothwell1 you’re right! Thanks for pointing it out. It was a little distraction…
@dibeos No problem. I appreciate the effort you guys put into the video (and making the PDF available, as usual).
You might be interested in seeing this related proof by Dr Barker:
"Differentiating x^{1/n} from First Principles without Power Series"
ruclips.net/video/-A7T8_5zhoc/видео.html
I was literally just thinking about this last night right before i slept... crazy coincidence that you guys uploaded a video on this exact topic the next day
Maybe it is not a coincidence... 👽👀
2:06 Finding Coincidences and patterns are what solidify our understanding-make those new neural connections 🎉
Something is off, at 9:00 rhe upper bound of the sum is changed from n to n+1.
Wonderful video 🙌
By far the trickiest part of this process is proving the binomial theorem with induction!!
my favorite proof of the power rule (or at least the one I have memorized) is MUCH shorter:
y = xⁿ
ln(y) = nln(x)
If we implicitly differentiate both sides we get:
y'/y = n(1/x)
Now just solve for y' and sub in y = xⁿ
y' = (n/x)y = (n/x)xⁿ = nx^(n-1)
Ok, I know using implicit differentiation and the derivative of the log is a little heavy handed but it leads to such a simple proof 😅
Great video!
@@graf_paper thanks! yeah, this proof is much shorter, but you do need to rely on the fact that one knows how to differentiate ln(X). I guess that’s the trade-off. Anyway, thanks for sharing. I didn’t really know hahah and yes, you’re right, I could title the video: how to prove the binomial theorem since most of the video is about that 😎
@@dibeos not a critique, I learn a lot from you both.
That's cool. That's very cool.
@@samueldeandrade8535 yeah, you definitely lost your dark powers! Thanks!
the result for rational exponents follows from implicit differentiation.
Very interesting video
I hope you can generalize the proof for n = any real number not only natural number
I think the approach taken here could potentially give some people, especially those new to these concepts, the impression that math/derivatives/proofs are always much more complicated than they need to be. Really, it's not too hard to show using just the distributivity of multiplication that (x+h)^n = x^n + hnx^(n-1) + other terms with powers of h of degree at least 2. The rest of the proof follows easily. I'm sure you're already aware of this. I'm just against the idea of overcomplicating simple stuff that a student might come across early into his learning journey. Concepts such as the binomial theorem can be introduced later in other contexts where they're more relevant.
It was great😀
@@User_2005st thanks! I’m glad you liked it 😊👍🏻
n belongs to real numbers