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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?
• 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.
• 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!
• what is e actually and what is its definition and how?thanx in advance
• does anyone know where i can find some good videos on e? or what the video is called here?
• Introduction to compound interest and e
• 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.
• 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.
• so is the derivative of 3e^x = 3e^x ?
• 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)
• Why 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
• 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.