Proof that the derivative of e^x is e^x.

Watch the next lesson:

Missed the previous lesson?

Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits.

About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We’ve also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.

For free. For everyone. Forever. #YouCanLearnAnything

Subscribe to KhanAcademy’s Differential Calculus channel:

Subscribe to KhanAcademy:

Nguồn: https://otoxetai.vn

Xem thêm bài viết khác: https://otoxetai.vn/xe

### Xem thêm Bài Viết:

- Công nghệ sạc thông minh có thể giúp giảm côn suất điện năng không?
- Bánh xe trên xe buýt | video cho trẻ em | Nursery Rhymes | Wheels On the Bus
- Cào Cào Địa Hình Leo Núi – Top Xe Hay
- Xe đua lắp ráp Tamiya cực ngầu và chạy nhanh ToyStation 317
- Giã nát 1 nắm rau rắm đắp lên mặt lỗ chân lông to như sẹo rộ cũng thu khít 100% da lại trắng hơn

Thanks❤❤

You can use the definition of a derivative to solve this which would make the proof seem more valid.

sir find[d/dx(x^e)]

"I think the school or the nation should take a national holiday and respond to this."

The last step is wrong

U r multiplying e^x on both sides but on the RHS u have to multiply it out of the derivative that is d/dy. e^x.1/e^x.that step is wrong

#impressed

Seems like someone enjoyed quite a few drinks before recording this video.

Your title is misleading somehow… d(e^x)/dx = e^x 😉

❤❤❤

so the video used d/dx (e^x) in a chain rule to prove that d/dx(e^x) = e^x. I don't think that would be considered valid proof since ur using the result with chain rule to prove the result.

So I spent an hour proving to myself that the derivative of lnx=1/x.. Then I was gonna try the derivative of e^x and I got lazy.. Now I wish I did it alone because it looks so easy.. But I have just one question.. Isn't d/dx(e^x) it's own term ? In a proof, you can't assume the right answer in it, you must prove the right answer. If we don't know the derivative of e^x is e^x yet, why are you saying that (e^x(d/dx(e^x)))/e^x=e^x. I'm only in algebra II, so I could really be wrong here.. But does anyone else feel the same ? I'm a bit confused..

Derivative of e^x is the y coordinate

Derivative of 1/2*x^2 is the x coordinate

I think they should be equally celebrated

I have a way of finding the derivative of e^x without using the chain rule or properties of natural logs.

d/dx(e^x)= lim(h->0) [e^(x+h)-e^x]/h = e^x. lim(h->0) [e^h-1]/h = e^x.lim(h->0) [1+h-1]/h=e^x

Proof for e^h=1+h :

Consider two functions "e^h" and "log_{1+h} {e^h^2.(1+h)}"

It is easy to see that both these functions approach 1 as h approaches 0.

So for values of h very very close to 0, we can say that:

e^h = log_{1+h} {e^h^2 * (1+h)}

= log_{1+h} {1+h} + log_{1+h} {e^h^2}

= 1 + h^2/log_e{1+h}

= 1 + h/[log_e{1+h}]/h

= 1+h/log_e [(1+h)^(1/h)] –(1)

Now for values of h close to 0, (1+h)^(1/h) ~ 2.71828… = e –(2)

So putting (2) in (1): e^h = 1+h/log_e{e} = 1+h

thank Youuuuuu

useless video

e^(iπ)+1=0

4:14 This is why I love Sal.

lets take the deridadef of well well la.. wer wurr

are you drunk sal?

I just mathgasmed

if d/dx lne(e^x)=1/e^x and d/dx e^x=1/e^x then

d/dx lne(e^x) = d/dx e^x

doesn't that mean that lne(e^x) = e^x

lne(e)*x=e^x

x=e^x

has anyone else noticed that Sal sometimes sounds drunk in his videos?

When I try and find the integral from 0-1 for this function, i run into the issue of receiving -infinity for 0 with the way i integrated it. So i figure im doing something wrong.

How do you solve for exponents such as negative square root of 'x' ( e^-(x^(1/2)) )

1/(e^x) is not being derived. It e^x is being derived, which means you're in the clear.

The poof goes like this:

1. Take the derivative of ln(e^x) by first using log rules to show ln(e^x) = x. d/dx (x) = 1.

2. Now that you know that, go through taking the derivative of ln(e^x) a second time. This time, you use the chain rule.

3. AFTER using chain rule, you get:

(1/e^x) * ( d/dx(e^x) ) = d/dx( ln(e^x) ) = 1 [From Step 1]

Or simply:

(1/e^x) * ( d/dx(e^x) ) = 1

Multiply both sides by e^x to get rid of the fraction, and you're done.

I understand the chain rule but how did you make the jump at 2:43

Also when you multiplied both sides by e^x, you can't because 1/(e^x) is being derived, or so I thought?

Two things I don't understand, I thought we were trying to prove d/dx(e^x), but in this video he starts by doing it as d/dx(ln(e^x)). I'm assuming ln(e^x) = e^x?

Next he's doing the chain rule to ln(e^x). First step is easy, taking the derivative of the inside which is e^x. Then the next step, which is supposed to be the exponent – 1 (n – 1), he gets 1/e^x instead. He got that from taking the derivative from lnx = 1/x… but I don't understand how that correlates.

I thought d/dx[e^x] = 42 🙂

Lol I'm stupid u guys are nerds

I guess this formula found by accident..

but because h cannot be 0 it cannot differentiate exactly to e^x, but i could be wrong :L :/

I'm pretty sure you went wrong somewhere here, you differentiated ln(e^x) to get 1, then differentiated it a different way to get 1/e^x, now assuming that makes sense you say that they must equal each other so 1/e^x=1 so 1=e^x you cant just scribble out the 1. a way to prove it is to say on the curve of e^x we can choose two points, p and p+h where h is the difference in the x axis between the two points so dy/dx((e^(p+h)-e^p)/h) as h gets closer to 0 it becomes effectively e^x

…derberdbebrerebdederebeberrrrebed.

E to the X!

Yes, well done. How is that relevant?