If they're teaching calculus, have the not taken advanced calculus? They SHOULD know the proof... I haven't taken that class yet, but that class should be "Calculus, but you have to prove everything you use."
You’ve looked at an extremely small sample of calculus teachers! Not only can calculus teachers prove the power rule, they can prove it in many different ways, including using the binomial theorem, or mathematical induction. For low values on n, there are nice geometrical visualizations.
I'm so glad the narrator mentions this has only proved the case for integers... positive integers to be specific because the binomial theorem only applies for positive integers. However to prove the rule is valid for all real numbers is a simple matter using logs and implicit differentiation. Recall the log property that ln(a^b)=b*ln(a)... So given y-x^n, take the (natural) log of both sides... ln(y)=ln(x^n)=n&ln(x) Now differentiate: (1/y)dy/dx=n/x ... Solve for dy/dx ... dy/dx= ny/x ...but y=x^n so, dy/dx=nx^n/x=nx^(n-1) QED.
I think the fastest way to explain it would be to just expand (x+h)^n. The first term, x^n cancels out with the second. The second term is n * x^(n-1) * h, and the h gets cancelled out by 1/h. The other terms will have an h to the power of 2 or greater, so they all tend to 0, and all you're left with is n * x^(n-1).
How so? I guess you could use Pascal's Triangle or the Binomial Theorem on (x+h)^n, but that would be a mess of coefficients, wouldn't it? (Though all but the first n term gets zeroed out anyway) The difference of powers formula seems like a clever step to avoid all that.
@@nilsh5027 I just used (x + h)^n = Σ_(k = 0)^n nCk x^(n - k)h^k, where that’s the sum from k = 0 to n, and nCk is the choose function (binomial coefficient/Pascal’s Triangle). Using this, any term with an h^2 or higher goes to 0 (since we divide by h, they all are order h^1 or greater) and we are left with x^n and nC1 x^(n-1). x^n is subtracted off by definition of the derivative and nC1 is identically n, leaving us with d/dx(x^n) = nx^(n-1)
@@mockman100k yeah, I looked it up before making that comment, but that's not the method the video used (which it seemed like OP was implying). I feel like the method in the video is a bit cleaner anyway, since you don't even have to think about the coefficients.
I have an old book with a similar explanation but I'm not getting it. Maybe you could give me a hand trying to solve this: f(x)=x^3. I proceed like this lim x->a (f(x)-f(a))/(x-a). Substitute (x^3-a^3)/(x-a) Then ((x-a)(x+a)^2)/(x-a) Cancel x-a and you're left with (x+a)^2 Substitute (x+x)^2=(2x)^2=4x^2 which isn't the expected result, 3x^2. Can you please point out what I'm doing wrong? Thank you so much!
Given a^(n-1) + a^(n-2)b + ... + ab^(n-2) + b^(n-1), notice that the powers of a go from n-1 to 0 (whereas b goes from 0 to n-1). That's the same as n...1 or 1...n which is n many terms.
the point of the derivative is to calculate the slope of the tangent at a point in the function wich represents the rate of change of a function in that point let point A be the point wich we want to find the slope of its tangent or in another words the point that we want to find its derivative lets pick a point nearby and call it B lets draw a line from point A to point B and calculate its slope slope = yb-ya/xb-xa okay well with this point B we can get an approximation to the slope of the tangent at point A this approximation will get better and better if we make point B approash A lets say that point XA = 1 we want point B to be soo close to XA so maybe B = 1,00000000000001 the difference between XA and XB is way smaller than that infact its infinitly small whats the difference between XA an XB ? its 0,00000000000000000001 so we say that XA = XA and XB = XA+H where H is infinitly small so lets calculate the slope of that now slope = y(a+h) - ya / xb-xa or = f(x+h) - f(x) / xb-xa we know that xb-xa = h cz its the infinitly smallnes so we just wite it as f(x+h) - f(x) /h and thats it um i hope this helps
I'm doing something very wrong, the rule for a^n-b^n isn't true when I work it out. I get 2 as a coefficient for AB when using the rule for (a^3-b^3) and it doesn't work for a^2-b^2 either. edit : Yeah I was doing it wrong. What I did was I used four terms for powers of 3 and 2 in the second bracket, where I should have used 3 and 2 terms in the second bracket respectively. very much my bad.
This proof explained without math: The power rule is just a handy thing to remember to make your life easier. If you do the math by hand, it will be exhausting, but you will get the answer in the end.
no teacher knows why the power rule is a thing, they just remember it. thanks so much for fulfilling my curiosity
I doubt that very much. If the teacher is teaching calculus especially.
If they're teaching calculus, have the not taken advanced calculus? They SHOULD know the proof... I haven't taken that class yet, but that class should be "Calculus, but you have to prove everything you use."
@@briendamathhatter816 It's called "Analysis". Pretty good stuff actually.
You’ve looked at an extremely small sample of calculus teachers! Not only can calculus teachers prove the power rule, they can prove it in many different ways, including using the binomial theorem, or mathematical induction. For low values on n, there are nice geometrical visualizations.
that was beautifully explained
I'm so glad the narrator mentions this has only proved the case for integers... positive integers to be specific because the binomial theorem only applies for positive integers. However to prove the rule is valid for all real numbers is a simple matter using logs and implicit differentiation. Recall the log property that ln(a^b)=b*ln(a)... So given y-x^n, take the (natural) log of both sides... ln(y)=ln(x^n)=n&ln(x) Now differentiate: (1/y)dy/dx=n/x ... Solve for dy/dx ... dy/dx= ny/x ...but y=x^n so, dy/dx=nx^n/x=nx^(n-1) QED.
thank you that drove me straight to the point
I think the fastest way to explain it would be to just expand (x+h)^n. The first term, x^n cancels out with the second. The second term is n * x^(n-1) * h, and the h gets cancelled out by 1/h. The other terms will have an h to the power of 2 or greater, so they all tend to 0, and all you're left with is n * x^(n-1).
They don't cancel.
You have x^n-1+x^n-1+...+x^n-1+x^n-1
Therefore you have n many x^n-1 and you can conclude that f'(x)=nx^n-1
I just realized that uses Pascals triangle
How so? I guess you could use Pascal's Triangle or the Binomial Theorem on (x+h)^n, but that would be a mess of coefficients, wouldn't it? (Though all but the first n term gets zeroed out anyway) The difference of powers formula seems like a clever step to avoid all that.
@@nilsh5027 I just used (x + h)^n = Σ_(k = 0)^n nCk x^(n - k)h^k, where that’s the sum from k = 0 to n, and nCk is the choose function (binomial coefficient/Pascal’s Triangle). Using this, any term with an h^2 or higher goes to 0 (since we divide by h, they all are order h^1 or greater) and we are left with x^n and nC1 x^(n-1). x^n is subtracted off by definition of the derivative and nC1 is identically n, leaving us with d/dx(x^n) = nx^(n-1)
@@mockman100k yeah, I looked it up before making that comment, but that's not the method the video used (which it seemed like OP was implying). I feel like the method in the video is a bit cleaner anyway, since you don't even have to think about the coefficients.
This is truly beautiful
THANK YOU TANK YOU THANK YOU THANK YOU......
Great explanation😀
great help, thank you
Awesome.
Sir please make more video like this regarding derivative,trigonometry.
and other vast problems
The most logical way to explain the power rule
Finally understand the formal proof of this, thanks!
Wait, how did you know that there would be n (x-1)s?
Hint for general power rule: chain rule, logarithmic differentiation
+Martin Nolin That won't give you the general power rule. That will give you the power rule for the set of n belongs to the rational numbers.
Beautiful
Binomial expansion of (x+h)^n = x^n+nx^(n-1)h+...+h^n
I have an old book with a similar explanation but I'm not getting it. Maybe you could give me a hand trying to solve this: f(x)=x^3.
I proceed like this lim x->a (f(x)-f(a))/(x-a).
Substitute
(x^3-a^3)/(x-a)
Then
((x-a)(x+a)^2)/(x-a)
Cancel x-a and you're left with
(x+a)^2
Substitute
(x+x)^2=(2x)^2=4x^2 which isn't the expected result, 3x^2.
Can you please point out what I'm doing wrong?
Thank you so much!
(x^3-x^a) does not factorise to (x-a)(x+a)^2. That’s the problem.
third root of x please
Thx alot mate
Mindblowing
How do you know at the end that you have an “n” amount of those X^n-1 functions?
Given a^(n-1) + a^(n-2)b + ... + ab^(n-2) + b^(n-1), notice that the powers of a go from n-1 to 0 (whereas b goes from 0 to n-1). That's the same as n...1 or 1...n which is n many terms.
@@nilsh5027thank you
🤩Appreciate
how do you know this works when n isnt a natural number???
Alex Sere im late AS FUCK lol i don't know if you found the answer but this can be proven by the implicit differentiation
N = a constant , that means N can be any number
@@RamsLiff no, he used the conjugate formula (a-b)(a^n+ ba^(n-1)...+b^n, which is only valid for POSITIVE INTEGER n
@@alexsere3061 use the binomial expansion formula (not nCr, i mean the one used for all real numbers)
@@alexsere3061 learn binomial expansion. It would make sense to you
you need to go into more detail about how all those x's cancel. you still have a lot of x^n-1 which didn't cancel.
Proof that 1+2 = 3.
Assume 1+2 =3
done.
where does (x+h)^n = something come from?
(x+h)^n = f(x+h)
the guy is doing f(x+h) - f(x)
if you dont understand please go learn derivatives, it comes from the definition
Alex Sere no thats not what im talking about. where did (x+h)^n = something come from. where did the SOMETHING come from. I understand it now
you understand it now? great!
Alex Sere :)
the point of the derivative is to calculate the slope of the tangent at a point in the function
wich represents the rate of change of a function in that point
let point A be the point wich we want to find the slope of its tangent or in another words the point that we want to find its derivative
lets pick a point nearby and call it B lets draw a line from point A to point B and calculate its slope slope = yb-ya/xb-xa okay well with this point B we can get an approximation to the slope of the tangent at point A this approximation will get better and better if we make point B approash A
lets say that point XA = 1 we want point B to be soo close to XA so maybe B = 1,00000000000001
the difference between XA and XB is way smaller than that infact its infinitly small whats the difference between XA an XB ? its 0,00000000000000000001 so we say that XA = XA and XB = XA+H where H is infinitly small
so lets calculate the slope of that now
slope = y(a+h) - ya / xb-xa or = f(x+h) - f(x) / xb-xa we know that xb-xa = h cz its the infinitly smallnes
so we just wite it as f(x+h) - f(x) /h
and thats it um i hope this helps
I'm doing something very wrong, the rule for a^n-b^n isn't true when I work it out. I get 2 as a coefficient for AB when using the rule for (a^3-b^3) and it doesn't work for a^2-b^2 either.
edit : Yeah I was doing it wrong. What I did was I used four terms for powers of 3 and 2 in the second bracket, where I should have used 3 and 2 terms in the second bracket respectively. very much my bad.
Wouldn't this proof be easier with (f(x)-f(x))/(x-a)?
This proof explained without math: The power rule is just a handy thing to remember to make your life easier. If you do the math by hand, it will be exhausting, but you will get the answer in the end.
This does not prove power rule for negative and fractional real number.
Why is that N word everywhere in maths?
n= num = number, plus X.y,z, a,b,c,u,v all have previous apps
In America, the phrase "n-word" refers to a racial slur. This comment was unintentionally comedic gold.
why does your lim look like this -> lım