So the reason we can cancel out the Δx is because when we expand the (x+n)^n (we're not caring for the actual values of the binomial coefficients here), every single term in this expansion will have a factor or Δx except the very first term which is x^n, we will be able to cancel out this first term x^n with the last thing -x^n which leaves us with a bunch of terms all having AT LEAST one Δx, thus we now can factor out one Δx and cancel it with the denominator, now the first term in our numerator will only be nx^(n-1) since it only had one Δx before factoring, the rest of the terms will now have a factor of at least one Δx except the first term, and when you now take the limit all terms goes to 0 except the first one (which once again is nx^(n-1), and that's the proof.
Amazing! I've been trying to create an intuitive proof of the power rule myself, and this is the only one I've found online.
Such a good proof! Thanks! Deserves more attention
So the reason we can cancel out the Δx is because when we expand the (x+n)^n (we're not caring for the actual values of the binomial coefficients here), every single term in this expansion will have a factor or Δx except the very first term which is x^n, we will be able to cancel out this first term x^n with the last thing -x^n which leaves us with a bunch of terms all having AT LEAST one Δx, thus we now can factor out one Δx and cancel it with the denominator, now the first term in our numerator will only be nx^(n-1) since it only had one Δx before factoring, the rest of the terms will now have a factor of at least one Δx except the first term, and when you now take the limit all terms goes to 0 except the first one (which once again is nx^(n-1), and that's the proof.
Thank you for such a positive contribution!
@@slcmathpc Why can't we write the delta x term as delta x to the power of n, as we do with x?
Helpful. Thanks.
very good 👍
Love it. Thx for the proof
Absolutely fantastic sir
Very clear :D