I thought this was a fun proof :D Product rule and Quotient rule proofs are coming soon. Keep flexin' those brain muscles! Instagram: / braingainzofficial
Minor technical details: you can't take the log of a negative number so this proof breaks down there. Also the power rule isn't actually true for every real n. d/dx x^0 = 0/x which is 0 for almost all x, but is undefined at x=0.
Well, i'd say that you can actually take the log of a negative number. Sure, the result would be complex, but since derivatives work the same for complex numbers, you could still say d/dx ln(x) = 1/x. So, although both x and f(x) can be negative, this proof seems correct.
@@debussy_69 The problem is really that ln(x^n) = nln(x) is not true when x is negative, not that we "can't" take the log of a negative number. So, this shows the power rule for all nonzero real n on the domain x>0.
Oh wow my comment got pinned lol. I'm honored! Also you can pretty easily extend this proof to the negatives by noticing that x^n is either odd or even depending on whether n is odd or even. The case where x=0 is trivial.
I'm almost offended that you don't have a lot more views. I hope you don't ever stop making videos, cause I'm starting on this math journey and videos like these save my life.
Epic Math Time sent me here! Apparently he, you, and I all agree that we don't like to just be given formulas to blindly use, haha! I remember trying to use the Binomial Theorem to prove the Product Rule a while back, and I found it so satisfying to have all those terms with an h-factor zero-out, leaving me with the expression I wanted!
Wow, I finished Calculus 3 last semester (thanks to God), and I was curious to know why the Power Rule works. I don't know the math behind the Binomial Theorem, but your first proof in this video seems so cool. It was SO satisfying to see that "h" in the denominator just go away.
Minor technical details: you can't take the log of a negative number so this proof breaks down there. Also the power rule isn't actually true for every real n. d/dx x^0 = 0/x which is 0 for almost all x, but is undefined at x=0.
Well, i'd say that you can actually take the log of a negative number. Sure, the result would be complex, but since derivatives work the same for complex numbers, you could still say d/dx ln(x) = 1/x. So, although both x and f(x) can be negative, this proof seems correct.
@@debussy_69 The problem is really that ln(x^n) = nln(x) is not true when x is negative, not that we "can't" take the log of a negative number.
So, this shows the power rule for all nonzero real n on the domain x>0.
@@EpicMathTime Oh okay, thanks for the correction 😅
@@EpicMathTime Damn. You're right.
Oh wow my comment got pinned lol. I'm honored! Also you can pretty easily extend this proof to the negatives by noticing that x^n is either odd or even depending on whether n is odd or even. The case where x=0 is trivial.
You just blew my mind with the alternative proof. Thank you and God bless.
I'm almost offended that you don't have a lot more views. I hope you don't ever stop making videos, cause I'm starting on this math journey and videos like these save my life.
I've seen the first method before, but I've never seen the second one. That was pretty cool.
Epic Math Time sent me here! Apparently he, you, and I all agree that we don't like to just be given formulas to blindly use, haha! I remember trying to use the Binomial Theorem to prove the Product Rule a while back, and I found it so satisfying to have all those terms with an h-factor zero-out, leaving me with the expression I wanted!
Nice video. I'm going to assign my calc 1 students to watch it this weekend.
Write x^n = x*x^(n-1) and use the product rule and induction.
As always great videos man. That Hendrix T-shirt is fire
thanks man!
You are really good at explaining! Really helped thanks a ton
Wow, I finished Calculus 3 last semester (thanks to God), and I was curious to know why the Power Rule works. I don't know the math behind the Binomial Theorem, but your first proof in this video seems so cool. It was SO satisfying to see that "h" in the denominator just go away.
Bundle of thanks 👍👍
Thank you so much! This was so interesting
Thank u sir🙏🏻 really helpful
Interesting!