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

Lesson 6: Proofs for the derivatives of eˣ and ln(x)

# Proofs of derivatives of ln(x) and eˣ

Doing both proofs in the same video to clarify any misconceptions that the original proof was "circular". Created by Sal Khan.

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• When using the chain rule in the proof that derivative os e^x=e^x, in , before proving that the statement is correct, I can't say that the derivativo od ln(e^x) = (e^x)(1/e^x). I'm assuming that de derivative o g(x) in the chain rule, in this case, e^x, is equal to e^x, that is just what I'm still trying to prove... I'm not 100% convinced. Is there any other way to prove that?
(1 vote)
• He didn't assert that
d/dx [ln(e^x)] = (e^x)(1/e^x). He asserted that
d/dx [ln(e^x)] = (d/dx [e^x]) (1/e^x). Do you see the difference?
• At : "the 1/x term has nothing to do with n", but x is part of the definition of n! I would say there is more needed here to satisfy us that the x term can be safely moved outside the lim(n->∞) term.

We define 1/n = Δx/x, and then claim n->∞ <=> Δx->0. Why are we free to ignore the other possibility: n->∞ <=> x->∞?
• As for the relation between x and n, x is just a constant. (It won’t change its value.)
If we rewrite 1/n=∆x/x as n=x/∆x, it is saying “x can be divided by ∆x n times” (n = changing rate between x and ∆x), or, you can see n as a function of ∆x, and x as a coefficient. Since x is a constant, you’ll find that as ∆x gets closer to 0, n would get closer to infinity. That’s how n is defined here.
Remember, I’m just talking about the relation between x and n. x originally, of course, is a variable.
Also, x originally came from ln(x), so it can’t be a non-positive number like Kaio pointed out.
• He mentions that in another video, he says that a definition of e is the lim as n->infinity of (1+1/n)^n.
Which video is he referencing here?
• "Introduction to compound interest and e" part 1, 2, ...
• could be explained in another video but: why is ln(ax), where a is any constant, the derivative still 1/x?
• ln(ax) = ln(a) + ln(x) (<-- Basic log rule!), where, as a is a constant, ln(a) is a constant. d(ln(a) + ln(x)) = d(ln(a)) + d(ln(x)), as ln(a) is constant, d(ln(a)) = 0.
Therefore, d(ln(ax)) = d(ln(x))
• @
I don't understand how he can just move the 1/(delta x) to the exponent.
The limit of Ln( 1 + (delta x)/x ) ^ ( 1 / (delta x) ) as delta x approaches 0 is 0 right?
• Can he switch from delta x aproaches 0 to n aprroaches infinity? Because,myes, if n approaches infinity, then delta x will approach zero. But also, if n approaches minus infinity, delta x will also approach zero... I haven't done the calculation, but i am not sure that n approaching minus infinity will give us e....
• n is defined as a non negative integer, so n cannot approach minus infinity. This is a flaw you must understand as it is fundamental to your future understanding of series and the limiting process.
• Is there a video that graphically displays these functions with regard to their limits? I'm having trouble visualizing the limits of n and delta-x with respect to their role in taking the derivative. Thanks
• what is the circularity Sal was talking about?
• Circularity is when you prove something using a definition that you haven't been proved. For example, if you are trying to prove ln x and e^x and you make the assumption that those derivatives are true in the proof, then you really haven't proven anything.
• I'm confused as to how e=limn-->infinity (1+1/n)^n it feels like it should be 1 as 1/n where n was infinity would be zero, i think?, and then 1^infinity would be one.
(1 vote)
• The 1/n term itself does tend towards 0, but it never equals 0. The base of your exponent is always larger than one. Also, since the exponent is getting much larger, the small part being added to the one causes it to become quite a bit larger than one. If you want to, try plugging some very large values of n into that formula on your calculator, and you can see it approaching 2.71828....
• Can we define e as a number so that the derivative of e^x is equal to the derivative of e^x? Then can we use that to find the derivative of ln x? Also, can we find the value of e using this method by assuming e^x to be a polynomial in the form a0+a1x+a2x^2... . Then by taking the derivative of that polynomial, we would get-
a0+a1x+a2x^2...=a1+2a2x+3a3x^2...
We would get-
a1=a0
a2=1/2a1
a3=1/6a1
And so on... .
Substituting x and a0 for 1, we get-
e=1+1+1/2+1/6...
This is approximately equal to 2.67, which is very close to the actual value of e.
Also by looking at the pattern we find that 1/ai=i!
Therefore we can express e as-
Autosum i=0 to infinite of 1/i!
Can this formula be used to find the value of e? Is this correct?