Differentiation a^x from first principles

Differentiating this function from first principles requires the use of the definition of the derivative and more importantly the limit definition of the natural log.

[@punditgi]

This channel is my prime source of great mathematics information! 🎉😊

[@xgx899]

you will regret it...

[@anonymouscheesepie3768]

For everyone curious about the limit definition of lnx as shown in this video, I was able to derive it myself by assuming two givens: 1. ∫x^(-1)dx [bounds: 1, x] = lnx 2. ∫x^ndx [bounds: 1, x] = [x^(n+1)]/(n+1) - 1/(n+1) = [x^(n+1) - 1]/(n+1) We can take the integral of x^-1 as a limit statement for some number t approaching -1: ∫x^(-1)dx = lim t->-1 ∫x^tdx Accordingly, given the bounds 1, x, the integral's result becomes lim t->-1 [x^(t+1) - 1]/(t+1) We want this as a limit of a variable going to zero, though. So we can introduce a new variable h such that as t goes to -1, h goes to 0: h = t + 1, transforming our limit statement thusly: lim h->0 (x^h - 1)/h. This limit statement is a representation of the integral of x^-1 from 1 to x. We also know that lnx = ∫x^-1dx [bounds: 1, x]. Hence, we can set the two equal: lnx = lim h->0 (x^h - 1)/h, thus proving the definition for the natural log function. There are other ways to do this, but I just thought this was the simplest and most intuitive way to go about it.

[@張謙-n3l]

Another way to approach the (a^h - 1)/h is to use the well-known fact that (e^h -1)/h is 1 a^h is e^ln(a^h), which by the logarithm rule, it is e^(h lna) so (a^h - 1)/h can be rewritten as (e^(h lna) - 1)/h, and we can multiple lna to the numerator and denominator to get lna (e^(h lna) - 1)/h lna, when h approach 0, y = h lna also approach 0, so we can now rewrite the whole thing as lna (e^y -1)/y which approach lna when y approach 0

[@xgx899]

if you use exponent, just write a^x=exp(xln a) and the derivation becomes trivial.

[@wannabeactuary01]

Wow! The limit definition of natural logarithm - how did I get a degree without it! Super clear explanation - thank you!

[@barryzeeberg3672]

7:14 - it would be better to show the derivation of the limit formula rather than to just write it down. Also, must confirm that the limit formula had not been derived by assuming the form for the derivative of a^x

[@solcarzemog5232]

I concur 100% I want to see the limit calculation.

[@windowsxpmemesandstufflol]

You just change a to e^lna and multiply and divide by lna then derive the limit of (e^u-1)/u [=1] where u=hlna approaching 0, then lna×1=1 and you're done

[@windowsxpmemesandstufflol]

oh btw use e=lim h->0 (1+h)^(1/h) [by h=1/x in the usual definition]

[@barryzeeberg3672]

@@windowsxpmemesandstufflol yes I totally agree that is the way to demonstrate it

[@nothingbutmathproofs7150]

You forgot to say that a>0! A very nice video is to derive all the log formulas using that simple integration formula you wrote for ln(x). BTW, this is the 2nd time I suggested that you make this video. The proofs are beautiful!

[@denisb_MG]

I thought the ln(a) limit is a consequence of the l'hospital's rule, which uses derivatives (circular definition).

[@RyanLewis-Johnson-wq6xs]

You’re awesome!

[@d-hat-vr2002]

Happy thanksgiving to you also! I really enjoy your videos and learn a lot! I want to share with you and everyone a beautiful construction I discovered during the pre-dawn hours of Thanksgiving day: Suppose we are given a square ABCD with a line segment JK. JK can be at any angle and location relative to the square. What is an algorithm to inscribe a smaller square PQRS in ABCD where 2 sides of PQRS are parallel to JK? Answer: 1. Draw lines AC and DB to locate the center of □ABCD; label the center point as O. 2. Extend JK (by approximately 3/2 times the JO distance) in both directions 3. Construct a perpendicular line to JK through O. Label the intersection point L (that is, OL ⊥ JK with L being somewhere on extended line JK) 4. Copy the length OL to locate points M and N along line JK, so that OL = LM = LN. 5. Draw a line through OM to intersect □ABCD in two points; call them P and R 6. Draw a line through ON to intersect □ABCD in two points; call them Q and S PQRS is the desired inscribed square! Proof of correctness available upon request.

[@chintamanimoghe1178]

Not from first principles, strictly speaking, since the limit has only been stated and not derived.

[@PrimeNewtons]

That's a definition for natural log. A definition is constructed not derived.

[@chintamanimoghe1178]

Oh, well. I was eagerly looking forward to learning how the result/definition is arrived at.

[@tr060]

​@@PrimeNewtons I really never heard of this kind of definition of the natural log. Is there a context or source you can give?

[@MichaelGrantPhD]

@@PrimeNewtons I too would be curious to see a source for that. It's not on Wikipedia or mathworld for instance. I was all geared up for tackling that limit somehow :-)

[@hectormartinpenapollastri8431]

But to show that is a definition of log you have to show it is equivalent to the standard one you provided (the integral one).

[@vasiledumitrescu9555]

It would be cool if you explained a an exercise where you have to the differential, rather than the derivative

[@KaivalyaChess]

sir you are great, i even emailed you problem!!

[@ericongwae6742]

From the video of differentiating e^x by first principle, it also entailed proving why (e^h--1)/h, as h goes to zero, =1. Using the same fashion, one can prove why (a^h-1)/h, as h goes to zero, is equal to ln a.

[@mikefochtman7164]

And notice just for completeness, if a is equal to e, then we have a^x*ln(e) and ln(e) is 1. So for the case where a is equal to e, the derivative of a^x equals a^x.

[@ben_adel3437]

I tried to prove e^x yesterday i wasn't able to find it but then i found it out by firstly grouping e^x(e^h-1)/h then i put e^x out the limit so i deal with it later then i realized that the limit is gonna equal 1 but i didn't know how to prove it i thought of the limit definition of e because i ended up with (1+1/h)^(h^2) which i wasn't able to do then i thought of changing the limit writing h as 1/h then the limit being from h->0 to h->infinity then I wrote e's definition which beautifully simplified imto 1+1/h And that left me with (1+1/h-1)/1/h the ones cancelled out and left 1/h/1/h and that's 1 i think you could've gone through this in your video it would've just been a lil different because of e^hlna but that's just not gonna matter when you make the h->1/hlna it'll still go to infinity

[@AmanuelSetegn]

Thanks for Applied maths

[@navaneeth.k.v]

Thanks

[@RyanLewis-Johnson-wq6xs]

D[a^x,x]=a^xLn(a)

[@Mediterranean81]

Nice

[@andrewparker8636]

I think you have to be very careful here about circular arguments. I'd first ask the question, what is your definition of a^x? You've made the assumption that such a function exists and it has the property that f(a+b)=f(a)f(b). You're limit definition of log also assumes the existence of the same function. The most common definition of a^x is actually in terms of e^x. I think you can technically define it as some sort of analytic continuation of a^q where q is a rational number, but that's just nightmarish. Anyway, my point is that I *think* trying to differentiate this from first principles is effectively circular.

[@Abby-hi4sf]

Will you please prove the derivation of as t aproaches zero ln(x)= (x^t -1)/t

[@tradicionalfuturismo]

Great video. Is it necessary for "h" to be positive in order to make the factoring? If you assume a positive "h" you have only a right hand limit, right?

[@PrimeNewtons]

h is always positive because it is a measure of distance from the point in question. Distance can be from the right or left.

[@tapwater76]

@@PrimeNewtons "magnitude "

[@GrifGrey]

How do you get that limit definition of the natural log?

[@sobolzeev]

I am afraid all the piece is one vicious circle. What is the definition of a^x? By the definition, a^x = exp(x ln(a) ).

[@peterjansen4826]

Next video, show WHY this is the definition of ln(x)?

[@penguinb4u]

This is the usage or formula from memory, not derivation from first principles.

[@Nickesponja]

Now prove that those two definitions of ln(x) are equivalent, and that, if x>0, ln(e^x)=e^ln(x)=x

[@Mediterranean81]

f(g(x)) = x (f and g are inverses) Let f : E-> F and g : F->E and x € E We have y = f(x) That means that g(y) = g(f(x)) Since g is an inverse function of f x = g(y) by definition So g(f(x)) = x g(x) = ln (x) f(x) = e^x g o f (x) = x f o g (x) = x

[@Technichian462]

Sigh… my high school math teachers get an overall failing grade. I’m 61 now. I never learned this, as it was always taught, from the text book, exactly as it was in the text book. Should you ask a question, they made you feel stupid as shit, and everyone laughed. I moved around a lot when I was a kid. The only thing I excelled at was the things I was interested in, like science, electronics, and computers. English, math, history for the most part, but evidently I know learned mot history than most Freshman in college. I self studied my way in electronics, so that by the time I was a junior in high school, I was occasionally correcting the teacher. But usually he would say, “no, we haven’t covered that yet. That will be in the Advanced electronics class, next year, if you qualify” he said with a big grin. I was automatically enrolled in his next years class. But back to the math. When you have “A+B*c” i was always looking at it literally. I had to make up values for them. So in all cases, A=1, B=2, C=3. So the answer was always 9. They never taught the variables as variables. I did not learn anything about variables until I started programming in Basic (again, self taught). So it would have been helpful, to have actual numbers instead of variables, so I could see what was actually going on. I’m autistic, actually, I have Asperger’s. Some things come natural to me. Like electrical/mechanical trouble shooting. But it wasn’t until I began failing High school math, that I learned, I had no idea how to do long division. I was always just estimating. And I got pretty good at it, but I believe that was because they kept using the same things over and over and I passed because they kept using the same things again and again. I would love to see this stuff using real world numbers. I’ve seen people put on a show, writing long ass formulas, get applauded, and me sitting there with a stupid look on my face saying, “ok, thats just the same thing showed in the class. How do you use that to figure out anything?” Room busts into laughter. This was the late 1970’s. In Indiana. Where they did not care if you learned anything that had nothing to do with farming. FFA, the only class I cheated in. Becaue I could really care less what kind of corn grew where, or what a jersey cow looked like.

[@الأستاذمحمد-ش3ع]

Hello, I did not understand well. Can you simplify the matter more or put another video that explains and clarifies more?

[@vasiledumitrescu9555]

What is it that you did not understand?

[@xgx899]

a^x= exp(xln a), whence the derivative is ln a*exp(xln a)=ln a*a^x. Things like this should not take more than 5 seconds.

[@leofigoboh1611]

This is not from first principles

[@xgx899]

@@leofigoboh1611 To the contrary, in advanced texts, log x is defined as int_1^x dt/t, e^x, as the inverse function, and a^x is DEFINED as exp(a ln x). This is done precisely to make the theory elegant and precise (with the usual equations easily obtained). Do not confuse elementary pseudo-historical presentation with the "first principles".

[@erichlf]

I still like y = a^x => ln y = x ln a => y'/y = ln a => y' = y ln a => y' = a^x ln a

[@dante9632]

Cant we just use l hospital rule for that limit?

[@itsphoenixingtime]

Would be circular reasoning. You'd need to do d/dh a^h which is what you're finding at the start.

[@xgx899]

formula for the ln x is not a "definition".

[@cycklist]

A reminder that only your country has this "thanksgiving" thing, and that 200 other countries exist.

[@anonymouscheesepie3768]

just let ppl enjoy their country's holidays, its not always just about what you or others celebrate

[@danielkantor3248]

@cycklist, Or, you could choose to not be offended. The world is not responsible for the choices you make.

[@ArchDudeify]

You are cool 😎

[@kornelviktor6985]

Happy thanksgiving

[@ThomasAllen-n2r]

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[@eccentricaste3232]

First principle is garbage water.😂

[@abhiroopkumar1064]

Why ..?

[@anestismoutafidis4575]

D={ α=ℝ\±e, 0} y'= x•α^(x-1)y'=xα^x/α^1; y"= (x-1)•xα^x-2 y"=x^2•α^(x-2)-xα^(x-2) y"=[(x^2•α^x-xα^x)/α^2] y"'=(x^2-x-2x+2)•[xα^(x-3)] y"'=x^3•α^(x-3) -3x^2•α^(x-3)+2xα^(x-3) y"'=x^3-3x^2+2x

[@jamesharmon4994]

Isn't there a "da" or "dx" in the answer?

[@benniepieters]

???