Proof of the Power Rule

Download the workbook and see how easy learning calculus can be.

[takyc]

no teacher knows why the power rule is a thing, they just remember it. thanks so much for fulfilling my curiosity

[barthennin]

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.

[mismis]

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).

[Lucas-zdhl]

I just realized that uses Pascals triangle

[bakayaro]

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

[maryamalkholi]

thank you that drove me straight to the point

[SherKhan]

Wait, how did you know that there would be n (x-1)s?

[alexsere]

how do you know this works when n isnt a natural number???

[martinnolin]

Hint for general power rule: chain rule, logarithmic differentiation

[lelcetz]

Proof that 1+2 = 3.
Assume 1+2 =3
done.
where does (x+h)^n = something come from?

[JSSTyger]

Binomial expansion of (x+h)^n = x^n+nx^(n-1)h+...+h^n

[Paulelele]

Finally understand the formal proof of this, thanks!

[Gattrecity]

The most logical way to explain the power rule

[krishmainali]

Awesome.
Sir please make more video like this regarding derivative, trigonometry.
and other vast problems

[dannyo]

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!

[cooldawg]

How do you know at the end that you have an “n” amount of those X^n-1 functions?

[kylefoley]

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.

[blakba]

Wouldn't this proof be easier with (f(x)-f(x))/(x-a)?

[mellmellody]

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.

[ThugginGame-sbdu]

I would've used the binomial series