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### Course: Class 12 math (India) > Unit 6

Lesson 6: Proofs for the derivatives of eˣ and ln(x)# Proof: d/dx(eˣ) = eˣ

Proof that the derivative of eˣ is eˣ. Created by Sal Khan.

## Want to join the conversation?

- Proving this makes absolutely no sense to me. Is it bad if I dont understand anything of this?(6 votes)
- I personally don't think this is a very good proof. There is a much simpler way to prove the derivative of eˣ from the definitions of e and of the derivative. Here is that proof:

e is defined as lim h→0 (1+h)^(1/h)

Thus, eʰ = lim h→0 (1+h)^(h/h)

Thus eʰ = lim h→0 (1+h)

We will come back to equality later, but for now let us recall the definition of a derivative:

lim h→0 [f(x+h) - f(x)] / h

Thus, by definition, the derivative of eˣ is:

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

d/dx (eˣ) = lim h→0 (1/h) [eˣ ∙ eʰ - eˣ ]

Factor out eˣ

d/dx (eˣ) = lim h→0 (1/h) eˣ [ eʰ - 1]

Since e^x does not contain h, it can be factored out of the limit completely:

d/dx (eˣ) = (eˣ) lim h→0 (1/h) [eʰ - 1]

But remember we previously established that

eʰ = lim h→0 (1+h)

So, let us substitute this into the derivative:

d/dx (eˣ) = (eˣ ) lim h→0 (1/h) [1+h - 1]

d/dx (eˣ) = (eˣ ) lim h→0 (1/h) [h]

d/dx (eˣ) = (eˣ ) lim h→0 1

d/dx (eˣ) = eˣ

This proof does not rely on any other derivative, nor the logarithm, but only on the definitions of e and the derivative and on very basic exponent and limit properties.

Now, I did skip over a little bit of proving that the two limits used are compatible, but it should be obvious that they are.(17 votes)

- What about d/dx of e^(-x)? please help me...(2 votes)
- This problem can also be solved using the chain rule.

Since the basic exponential function (using e as the base) is its own derivative,

d/dx of e^(-x) = e^(-x) * d/dx of (-x) = e^(-x) * (-1) = -e^(-x).

Have a blessed, wonderful day!(3 votes)

- what is e actually and what is its definition and how?thanx in advance(2 votes)
- does anyone know where i can find some good videos on e? or what the video is called here?(2 votes)
- Introduction to compound interest and e(2 votes)

- Interesting although technically a "proof" uses the definition. ie lim h-->0 f(x+h)-f(x)/h. Can you prove it that way?(1 vote)
- d/dx (e^x)= lim h->0 [e^(x+h)-e^x]/h = lim h->0 [(e^x)*(e^h) - e^x]/h = e^x*lim h->0 [(e^h-1)]/h

On the side set e^h-1=u. That means as h->0 u->0 as well. This would mean that e^h=u+1 and h=ln(u+1). So back to where we left off in the original function we have: e^x*lim u->0 u/[ln(u+1)]. This is an indeterminate form so you technically could use L'hospital's rule if you know what that is. If you don't then you multiply the top and the bottom by 1/u. This leaves us with e^x lim u->0 1/[ln(u+1)^1/u]. The trick in the denominator is to express the 1/x as an exponent of the logarithm as opposed to putting it just plain and simple in the denominator. The lim u->0 (1+u)^1/u = e. That is an identity which sal proved already. This means that we have e^x*lim u->0 1/(ln e)=1/1=1. Therefore we conclude that d/dx(e^x)=e^x.(4 votes)

- You have to use the chain rule to find the derivative of e^-x. The derivative of e raised to [something] is e^[something] times the derivative of [something]. Here, [something] is -x, so the derivative is e^-x times -1, which is the derivative of -x. So the answer is just -e^-x.(2 votes)

- so is the derivative of 3e^x = 3e^x ?(2 votes)
- That is correct, but only because the derivative of x=1.

Technically, d/dx 3e^x = (d/dx x) * 3e^x = 1 * 3e^x = 3e^x(1 vote)

- 2:05Why doesn't he use chain and product rule for xlne? Am guessing because this way we are proving a proof with itself which isn't much of a proof at all. Also, after this time, where is the derivation? He only took the natural logarithm of both sides and he got x, so dy/dx=x, so what is the slope? At 7 it's 7? This is rather mind boggling for me. Did he derive the x and got 1? If a derivative function is x,why are we looking at the slope of a slope/derived function and saying it's 1? xD Basically has the function been derived and it's x or did he simplify the function and then derive it to get 1? Basically, am not seeing any derivation being done here, just simplification, if we didn't derive it and we just simplified it, then how can it be x? Isn't x supposed to be derived? Sorry for writing the question over and over but I hope you get what am asking. I've learned logarithmic differentiation in the previous section and the first row of d/dx makes no sense to me because aren't we supposed to take the natural logarithm of both sides and then simplify, use log properties and afterwards derive?(1 vote)
- He does the derivative of ln(e^x) two different ways. First (in pink) he uses the rules of logarithmsto rewrite ln(e^x) as xln(e), but ln(e) = 1, so xln(e) = x. At this point *he still has not done any derivatives.*

So we have (in pink): d/dx(ln(e^x)) = d/dx(x) = 1.

Then, in light green, he finds the derivative of exactly the same function using the chain rule:

d/dx(ln(u)) = (du/dx)*(1/u), where in this case u = e^x.

So we have (in light green): d/dx(ln(e^x)) = d/dx(e^x)*(1/e^x)

Finally, he sets the pink equal to the light green:

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

which means d/dx(e^x) = e^x(3 votes)

- Why not say the derivative of e6x is x*e^x-1. [ because of that Power Formula that derivative of x^n is nx^n-1] .(1 vote)
- if derivative of ln(x) =(1/x) then shouldn't the derivative of ln(e^x) somehow be equal to (1/(e^x)) AND NOT 1(1 vote)
- No; to take the derivative of ln(e^x), you need to apply the Chain Rule, which you'll encounter later if you haven't already. By the Chain Rule,
`d/dx( ln(e^x) ) = [1/(e^x)] * e^x = 1`

. You can check out the Chain Rule videos for more information; I think Sal even has a video showing a proof of the Chain Rule, unless he's gotten rid of it recently.(3 votes)

## Video transcript

Let's prove with the derivative
of e to the x's, and I think that this is one of the most
amazing things, depending on how you view it about either
calculus or math or the universe. Well we're essentially going to
prove-- I've already told you before that the derivative of e
to the x is equal to e to the x, which is amazing. The slope at any point of that
line is equal to the x value-- is equal to the function at
that point, not the x value. The slope at any
point is equal to e. That is mind boggling. And that also means that the
second derivative at any point is equal to the function of
that value or the third derivative, or the infinite
derivative, and that never ceases to amaze me. But anyway back to work. So how are we going
to prove this? Well we already proved-- I
actually just did it right before starting this video--
that the derivative-- and some people actually call this
the definition of e. They go the other way around. They say there is some number
for which this is true, and we call that number e. So it could almost be viewed as
a little bit circular, but be we said that e is equal to the
limit as n approaches infinity of 1 over 1 plus n to the end. And then using this we actually
proved that derivative of ln of x is equal to 1/x. The derivative of log base
e of x is equal to 1/x. So now that we prove this out,
let's use this to prove this. Let me keep switching colors
to keep it interesting. Let's take the derivative
of ln of e to the x. This is almost trivial. This is equal to the logarithm
of a to the b is equal to b times the logarithm of a,
so this is equal to the derivative of x ln of e. And this is just saying e to
what power is equal to e. Well, to the first
power, right? So this just equals the
derivative of x, which we have shown as equal to 1. I think we have shown it,
hopefully we've shown it. If we haven't, that's actually
a very easy one to prove. OK fair enough. We did that. But let's do this another way. Let's use the chain rule. So what doe the chain rule say? If we have f of g of x, where
we have one function embedded in another one, the chain rule
say we take the derivative of the inside function, so
d/dx of e to the x. And then we take the derivative
of the outside function or the derivative of the outside
function with respect to the inner function. You can almost
view it that way. So the derivative ln
of e to the x with respect to e to the x. I know that's a
little confusing. You could have written a d e to
the x down over here, but I think you know the
chain rule by now. That is equal to 1
over e to the x. And that just comes from this. But instead of an x,
we have e to the x. So this is just a chain rule. Well what else do we know? We know that this is equal to
this, and we also know that this is equal to this. So this must be equal to this. So this must be equal to 1. Well let's just multiply
both sides of this equation by e to the x. We get on the left hand
side, we're just left with this expression. The derivative of e to the x
times- we're multiplying both sides by e to the x, times
e to the x over e to the x. I just chose to put the e
to the x on this term, is equal to e to the x. This is 1. Scratch it out. We're done. That might not have been
completely satisfying for you, but it works. The derivative of e to the
x is equal to e to the x. I think the school or the
nation should take a national holiday or something,
and people should just ponder this, because it
really is fascinating. But then actually this will
lead us to I would say even more dramatic results in the
not too far off future. Anyway, I'll see in
the next video.