Proof: d/dx(e^x) = e^x

[Michel0555]

How do you integrate 5*4^(3x+2)

[theoriginalcharlz18]

@Michel0555 rem the rule for differentiating a constant raised to a fn, say a^x that is a^x *lna. do the opp for the integrating part of it ie. a^x=a^x /lna,so in ur case it is 5*4^(3x+2) /ln(5*4)........

[Fr33Styla00]

What is the derivative of x/e^x ?

[stuntyannick2]

@Fr33Styla00 f'x = -(x-1)*e^-x

[theoriginalcharlz18]

@stuntyannick2 Check ur ans bcuz that looks like u have an x-1 inside the braces which is not correct,rem d/dx of x is 1 in this case -1 so u just multiply this by x=-x and multiply by e^x thus you have -xe^-x or -x/e^x.

[theoriginalcharlz18]

@Fr33Styla00 -xe^-x

[imperfectlego]

@jcarnt I don't see anywhere where he does that...

[TryptamineTea]

@imperfectlego Around 2:50. When he uses the chain rule he takes the derivative of the inside function which is e^x.

[jcarnt]

Really? You proved the derivative of e^x by taking the derivative of e^x?

Don't say you didn't because you took it in respect to e^x because that's not a variable. 

[GFauxPas]

@PCGamerPortal Be careful, the definition you're thinking of is (1 to x) ∫ (1/t )dt ≡ ln(x) which is a definite integral, not the antiderivative.

[TeenyOnesNursery]

ok, so i know what the derivative of e^x is, but how do i find the tangent line to xe^x at x=0? 

[GFauxPas]

@TeenyOnesNursery the first step is to find the slope of your mystery line. Do you know how to find the slope?

[CobyB162]

National e Day!

[drrobertoboogie97]

The same can be done with Euler's formula by assuming x:=iy and differentiating Euler's formula by x.

[asskickerone]

i love the way you love calculus. i agree. we should have a holiday to ponder this.

[efcdk92]

what about dy/dx of (2x^2-5x+2)e^-x ???

[drrobertoboogie97]

@efcdk92 (-2x^2+9x-7)e^-x

[Forman6]

e= (1+h^-1)^h as h approaches infinity thus common sense says e= (1+h)^(1/h) as h approaches zero. raise by power of h on both sides and you get e^h=1+h as h approaches zero which is one. now try (e^(x+h)-e^x)/h which e^(x+h)=e^x*e^h so you can write as (e^x* e^h)-e^x and you can take out e^x to put as e^x (e^h-1)/h but wait! e^h = 1+h as proven before! so e^x *(h+1)-1/h which is e^x h/h and finally limit of h approaches zero of ( e^x)*h/h? h/h =1 so you get e^x.

[BarbierNicholas]

@Forman6 hmm I see what you are saying...nice proof. Yet mine still works. It does rely on the prior assertion/proof (see another video by khanacademy-proof of derivative of ln(x)). Basically I am using the knowledge that d/dx(ln(x))=1/x (which by the way is almost the techincal definition of e. e's most basic definition is that the area under 1/x from 1 to e=1--the other definitions...i.e. lim (1+1/n)^n are all derived from this definition) It is still a rigorous proof just (f(x+h)-f(x))/h

[BarbierNicholas]

@Forman6 hmm I see what you are saying...nice proof. Yet mine still works. It does rely on the prior assertion/proof (see another video by khanacademy-proof of derivative of ln(x)). Basically I am using the knowledge that d/dx(ln(x))=1/x (which by the way is almost the techincal definition of e. e's most basic definition is that the area under 1/x from 1 to e=1--the other definitions...i.e. lim (1+1/n)^n are all derived from this definition) It is still a rigorous proof just not (f(x+h)-f(x))/h

[Forman6]

@BarbierNicholas yeah I liked yours.

its much simpler!

and we did it in class today!

Aren't proofs just gorgeous?

thats why i'm a math major!

and sorry if i mispell so much but i not english major anyways.



[BarbierNicholas]

@Forman6 Cool. that's awesome--being a math major. Where do you go?

I love proofs so much. I guess it is the fact that when something is proved (by definition) it is ABSOLUTELY true.

Unlike science where you match a model to fit observation, with math the numbers are self-consistent. The beauty found in mathematics (even though we invented it!) transcends anything i have ever seen.

[cepson]

I am not a mathematician. Still, there is something unsatisfying about this proof. I'm certainly not contesting it, it's just that if I were Euhler or Newton or whoever, wouldn't I have figured out what d/dx(e^x) was before I went to work on d/dx(ln(e^x))?

Eh, maybe not. Anyway, your videos are, as always, incredibly useful and well-explained. Thanks for doing this.

[Forman6]

@cepson yeah.

its not a real hardcore proof.

the real proof is by limit definition.

as h approaches 0 of (e^(x+h)-e^x)/h

[Forman6]

@cepson correction i meant by definition of a derivative

[BarbierNicholas]

I like my own method. Using the already proved definition that d/dx(ln(x))=1/x you can show:

e^x=y, x=lny

d/dy(x)=d/dy(lny)

dx/dy=1/y

dy/dx=y, using orginal definition that e^x=y you substitute

and voila! dy/dx=e^x

[Forman6]

@BarbierNicholas thats not a proof.

its just a nice way of showing e^x is e^x when you already know the derivative.

the real proof you are unknown of solution so real proof is by limit definition.

(e^(x+h) - e^x)/h and the limit of h approaches infinity and you get your derivative of e^x.

which is e^x.

try it.

[BlackDK18]

There was a math party, and all the functions were invited; the exponential function was alone in the corner, so the sinx came to it and said:

- Why are you alone? Come and integrate with us.

The e^x said:

- What for? It would be the same, anyway...

[magnetmixer]

u stutter a little bit...

[MarvelsofaLifetime]

What would the function e^x look like on a graph?

[Hellsslave666]

@MarvelsofaLifetime

It'a exponent funktion. Looks verry roughly like 2^x.

Try searching in google for a better result.

[Sephiro12]

@khanacademy I am doing calculus hw now and have been looking online for like an hour to find a description of why the derivative of e^x is e^x (and more importantly why it seems to be more specifically like if y=e^x, y'=((d/dx)x)(e^x)). Your video has been helpful, but is confusing at one point... when you multiply both sides by e^x, you get (d/dx)(e^x) = e^x. Why do you not multiply the (d/dx)(e^x) by e^x and get e^x(d/dx)(e^x) = e^x? Without doing this you get a nice proof but like.. huh?

[fewoifhiu3hfklew]

You can on forgetting to use the constants in your videos... shouldn't the question be "d/dx(ae^(bx)) = ?"

[prubeyou1]

My apologies to Sal. I posted a complaint about an error in his talk on the derivative of e^x (the first video). I misread what he put on the board (a product as an equality). My bad! You were right!

[dannyboy12357]

i like toprove it doing the easy way with a power series

[thomasking55]

e^(-x) worries me

[Svenskivsk]

ah new comment system sorry for the triple post

[Svenskivsk]

sorry 2/7

[Svenskivsk]

sorry 2/7

[Svenskivsk]

sorry 2/7

[Svenskivsk]

2/17 of every year is now e-day

[agricolaterrae]

You mentioned taking a holdiay to ponder the magic of "e". I think there already is one: Feburary 7. (2.7)

It's kind of like pi day (March 14th or 3.14)

[Johnedw2]

Great video Sal btw!

[Johnedw2]

It's a little misleading that e^x dervitive is e^x, it is true, but if we expand it a little bit, d/dx(n^x) = ln(n) * n^x where if n = e then ln(e) = 1 so ln(e) * e^x = 1 * e^x. The reality of exponential functions is that they are growing very rapidly. Another interesting reality is that d/dx(n^x) / n^x = ln(n)

[Johnedw2]

You can verify this by going to google and using the literal method of diffrentiation ((3^(2.000001) - 3^(2.0000001))/ (3.000001 - 3.0000001))/(3^2) will yield you ln(3)

You can use the same method to prove that d/dx(ln(x)) = 1/x, and an interesting side affect is that d/dx((d/dx(n^x)/n^x)) = 1/x

[RainyFeb18th]

My teacher gave me e^ax :( stuck

[gremlinextreme101]

you can do it by definition of the derivative. solve to

e^x * lim (e^h-1)/h and then using the series expansion of e^h we can see the limit goes to 1. No circular argument by using the series expansion as it can be found by an infinite binomial expansion of it's definition of

e^x = (lim n->inf) (1+x/n)^n

[woodenjaw]

Sal redid this video and reposted under the title "Proofs of Derivatives of Ln(x) and e^x" - High quality

[madscirat]

Sal, I luv yu dearly, but for the first time I must disagree. At ~4:30 you use d/dx e^x=e^x as part of your proof that d/dx e^x= e^x. You can't have your thesis as part of your proof, its circular. So the statement might be right but the route you used is technically fallacious. Perhaps I am mistaken, if so plz contact me. I would also be interested if there's a proof absent the circularity. If not the statement might be improvable which only serves to make it more fascinating!

[Elorviat]

Yes, he did use e^x in the proof, but he did NOT use d/dx e^x=e^x in the proof. That was the end result, which is what is supposed to happen.

[madscirat]

He does it when he uses the chain rule, derivitive of the inside is e^x which becomes e^x.... I think. Plz correct me if Im misreading, Id like to get this proof.

[Elorviat]

No, the derivative on the inside was rendered as d/dx e^x, showing that the derivative had not yet been proven. So, he let that remain. He then completed the chain rule by multiplying d/dx e^x by the derivative of ln (x). We know that d/dx ln(e^x)=1 because ln(e^x)=x and d/dx = 1. So we end up with (1/e^x)(d/dx e^x)=1. Multiply both sides by e^x and there you have it.

[ichinarukurumaki]

haha dude, 4:30 is when he writes his end result if u look about 30=40 seconds earlier and you will see.. d/dx(e^x*(1/(e^x)=1 ... this statement he clearly derived from his last video which stated that... d/dx(lnx)=1/x ... then he simply used the chain rule.. there was no rewriting and definitely no stating and assuming that d/dx(e^x) is equal to e^x and if that is not enough for you then u can go look at his newer video of the proof that he made because of all the hate and claims of circularity

[trevorbark]

love the begging... Mr. Stutter

[S1CKDRIFT3R]

but you have to use the fact of that the derivative of e^x is e^x to prove that the derivative of lnx is 1/x so how can you use the proof of the derivative of lnx to prove the derivative of e^x

[ww1313]

maybe february 7th can be 'e' day

[cressilia]

It was easy to me.

[cressilia]

and I'm in 6th grade.

[Chippy7070]

I never said it was a hard proof, just a bad one. No wonder you're in the 6th grade.....you should go back to 2nd grade and learn to read

[thebiggerhammer]

I finally understood the one in red once i remembered that it has to be equal to something...namely "y". It should say d/dx[y]=d/dx[e^x] then when you take the nat. log of both sides it becomes d/dx[Ln(y)]=d/dx[Ln(e^x)] Then: (1/y)dy/dx=d/dx [x] Then the right side becomes 1 and then multiply both sides by Y... Y=e^x therefore d/dx[e^x]=e^x

[tampiwa]

(de^x/dx) *(1/e^x) not d(e^x * 1\e^x)\dx mate!

[cwldoc]

That having been said, I enjoyed the video, which was very clear, well done, and shows how "it all fits together."

[cwldoc]

If one defines e as

lim n-> infinity [(1 + 1/n)^n], then in order for the assertion

[(e^x)'=e^x] to be defined, one must first prove the existence of

lim n-> infinity [(1 + 1/n)^n].

But if one proves the existence of this limit, for example by proving it is equal to

1 + 1 + 1/2! + 1/3! +. . .,

then the power series for e^x follows directly (at least for the real case); then you get the derivative of e^x from the power series, making the proof in the video redundant.

[cwldoc]

I believe there may be a problem with this proof. You are assuming the existence of

lim n-> infinity [(1 + 1/n)^n].

Was this proved in another of your lectures? (Of course you cannot use l'hopital's rule toward this end because that would necessitate knowing the derivative of e^x in the first place!)

[zsyed12345678]

And also, even if you assume the existence of the limit, in order to interchange the power function and the limit, you must prove that the latter is continuous. I agree that this would be overkill and should be hand-waived in first year calculus classes anyway :-) I liked the presentation too, very well done.

[Dirty420South]

Awesome ^_^

[pinkcap]

what is the derivative of e^lnx :S ? i have an exam in 2 dayz and its coming!

[khanacademy]

e^(ln x) = x (although you have to restrict the domain to x>0 since you can't take the ln of 0 or a negative number). the derivative of x is just 1.

[pinkcap]

oh now thats smthings i didnt know!i thought wer suppose to solve it like how we solve other composite functions.thnx alot for the help,i appreciate it.

[IndiaJacob]

@khanacademy wont we have to use the chain rule in this case?????

d/dx e^(ln x) = e^(ln x) * 1/x ..............please answer. i have the exam in 3 days and im totally depending on your videos cuz this is the first time i understood the concept. Thanks a lot :)

[RobertoEmilioRomero]

@pinkcap De Derrivative of e^Lnx with respect to x is 1.

[Patsan120]

the first 5 seconds were epic

[MelloMushroom91]

lol

[vonnegut85]

At the step when you used the chain rule, it was assumed that the derivative of e^x exists (when you wrote d/dx e^x). How do you show the derivative of e^x exists?

[clerlic]

It's a continuous function, therefore it should have a derivative (although there are rare exceptions).

[prcrazy12345]

i now this , i have problems taking out this derivatives e^2x=??? or e^2x^2=??? i know this are harder ones, but they came in my first 2 exams and its coming in my final, and i dont know how to do this.

[sikory]

if I can say something about it i would suggest with the first rule to rewrite e^2x as (e^x)*(e^x) and then use the product rule.

For the other problem (e^2x)^2=e^4x wich makes it also pretty straightforward

[sikory]

if it's e^((2x)^2) then using the chainrule is a simple way since the chainrule can be applied to pretty much anything to simplify it

[smartmusicfreak]

One of my favorite methods to show that d/dx(e^x)=e^x is showing the derivative of the sum of x^n/n! as n ranges from 0 to infinity.

In other words d/dx(1 + x + x^2/2! + x^3/3!...) = 1 + x + x^2/2! + x^3/3!...

[khanacademy]

good idea. Might make a video on that now that we've covered the maclaurin series of e^x (the series you cite)

[sikory]

However that would be a circular argument, since to construct the power series, you have already assumed that the derivative of e^x=e^x

[alisaffah]

@khanacademy

please can you give me the name of the software you used to plot the graphs (taylor polynomial lecture) thanks a lot

[Waranle]

Could please tell me what screep capture you use, Thank you