Given x = Ln[E^x] How do you derive the formula d/dx E^x = E^x with d/dx Ln[x] = 1/x ?

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Given x = Ln[E^x] How do you derive the formula d/dx E^x = E^x with d/dx Ln[x] = 1/x ?

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\[x=\ln(e^x)\] this?
oh yeah sorry, on my system all Logs are to the base E
I know based on one rule if \[f(x) = \ln x \] then the derivative is \[f'(x) = \frac{1}{x}\]

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that's a theorem Theorem (The Derivative of the Natural Logarithm Function
\[\frac{d}{dx} e^x =e^x\] The derivative of \[e^x\]with respect to x is equal to e^x
because the derivative of x is just 1 and that's normally not written down
an example find the derivative of \[e^{2x}\] let's just take 2x into consideration. The derivative of 2x is 2. So that 2 goes in the front \[2e^{2x}\]
is it related the the fact that E is a constant?
errr...not exactly
this is calculus ii related. there is a bunch of log and exponential derivatives out there
hmmm if I use the natural log definition on this...I think something should cancel out
I have a big list of them in my notes.. but this problem looks like it wants me to prove why E^x is the derivative of E^x
perhaps I need to throw the definition of the derivative at it?
wait to prove why e^x is the derivative of e^x. I think I typed it already
no no no no definition of the derivative isn't related.
I notice the derivative of 2^x is coming up as 2^x Log[2]
er Ln[2]
that's a different rule though
so I guess if c is a constant and \[\frac{ d }{ dx } c^x = c^x Ln[c]\] then \[\frac{ d }{ dx } E^x = E^x Ln[E]\] and if Ln[E] = 1 then \[\frac{ d }{ dx } E^x = E^x \]
ln e is 1? sorry I haven't done proofs in my CAlculus class....so it's a bit iffy for me
yeah I might be wrong on that.
x = Log[E^x]
Ln[E] = 1 is true
hughfuve - your calculator only does natural logs? (Base 'e' logs?)
yes sorry.. I keep forgetting that.
I'll try to use Ln[] in future
Heck, I wrote an online log calculator that will do logarithms of ANY base http://www.1728.org/logrithm.htm
cool stuff
gee thanks :-)
is a proof required?
I'm still kind of new when it comes to proofs. I know some, but not all... but I know computation :)
So, what needs to be proven?
derivation? at least that's on the op's question
so, only prove the derivative of e^x and Lnx? including Ln(e^x).
The problem reads... Start with the formulas \[x = Ln[E^x]\] and \[\frac{d}{dx} Log[x] = \frac{1}{x} \] and derive the formula \[\frac{d}{dx} e^x = e^x \]
and Im assuming that means, they want some kind of proof
chain rule? I mean.. \[\ln(e^x) = \frac{1}{e^{x}} e^x\]
but that e^x cancels out... x.x sorry at least I tried
yeah that's what I got.. I dont know what the hell they're asking for here.
derive just means take the derivative.. the question didn't say show that or prove or disprove.
I'm confused.
how do we derive the formula, when we just know it to be true?
If you need to prove the properties, I can show you how to prove them, but is that what the excercise requires?
sorry.. please remember that Log is base E
Im not sure... it says to start with x = Ln[E^x] and d/dx Ln[x] = 1/x and derive the formula d/dx e^x = e^x I'm assuming they want me to use those first two functions to arrive at the conclusion of the derived formula, through some kind of manipulation.
So, prove the derivatives and apply it to the initial function?
maybe it's one of those questions where you start from the left side of the problem.. apply theorems and rules and have that result match the right side of the problem
taking the derivative of ln x is 1/x
acutally 1/x (1) because the derivative of x is 1 but no one writes that these days
\[\ln x \rightarrow \frac {1}{x} (1) \rightarrow \frac{1}{x}\]
Maybe they want me to use logarithmic differentiation. If \[g[x] = Ln[f[x]]\] Then \[f'[x] = f[x] g'[x] \]
then by the chain rule \[\frac{d}{dx} Log[f[x]] = \frac{f'[x]}{f[x]}\]
which is why the derivative of Log[x] = 1/x because the derivative of x = 1
Well... That would be an incomplete proof, the right ways is using the very definition of derivative, so, therefore: \[f:fx)=lnx\] \[f:f(x) \rightarrow f':f'(x) <=>\lim_{x \rightarrow a}\frac{ \ln(x)-\ln(a) }{ x-a }\] So, all we have to do is solve the indetermination of the limit stated above (which is why you have to pretty much be skilled with limits before you do derivatives). So, let's take it and simplify, to see if we can solve that indetermination: \[f'(x)=\lim_{x \rightarrow a}\frac{ \ln(x)-\ln(a) }{ x-a } \rightarrow f'(x)=\frac{ \ln \frac{ x }{ a } }{ x-a }\] Now, there is an interesting equivalent I have to explain before continuing: \[Ln f(x) = f(x)-1\] \[f(x)\rightarrow 1\] So, it says the folowing, "The logarithm of any function is equivalent to the function inside the logarithm sustracted 1 , as long as the function tends to 1". So, returning to the proof: \[f'(x)=\lim_{x \rightarrow a}\frac{ \ln \frac{ x }{ a } }{ x-a }\] We can clearly see that x/a tends to 1, because x->a. therefore, by equivalence: \[f'(x)=\lim_{x \rightarrow a}\frac{ \frac{ x }{ a }-1 }{ x-a }\] \[f'(x)=\lim_{x \rightarrow a}\frac{ \frac{ x-a }{ a } }{ x-a }\rightarrow f'(x)=\lim_{x \rightarrow a} \frac{ x-a }{ a(x-a) }\] \[f'(x)=\lim_{x \rightarrow a}\frac{ 1 }{ a }=\frac{ 1 }{ a }\] \[=> f'(x)=\frac{ 1 }{ a }\] And that's the derivative of any given point in the function Lnx. a generalization would be: \[f:f(x)=lnx \rightarrow f':f'(x)=\frac{ 1 }{ x }\]
wow thats wicked.. a limit proof. I had cooked up this. GIVEN \[x = Ln[E^x]\] And \[\frac{d}{dx} x = 1 \] Then \[\frac{d}{dx} Ln[E^x] = 1\] GIVEN \[\frac{d}{dx} Ln[x] = \frac{1}{x}\] Then \[\frac{d}{dx} Ln[E^x] = \frac{1}{E^x} * \frac{d}{dx}E^x= 1\] by the chain rule \[\frac{d}{dx} Ln[E^x] * E^2 = \frac{1}{E^x} * \frac{d}{dx}E^x * E^2= 1 *E^2\] \[\frac{d}{dx} Ln[E^x] * E^2 = \frac{d}{dx}E^x = E^2\] Therefore the derivative of E^2 by this equation must be E^2
The same is applied for the derivative of E^x, we will state it, then apply the definition to it: \[f:f(x)=e^x\] \[f:f(x) \rightarrow f':f'(a)<=>\lim_{x \rightarrow a}\frac{ e^x-e^a }{ x-a }\] Yet again, we have to solve the indetermination stated by the limit of the definition: So, let's play around with it so to see if we can remove the indetermination: \[f'(a)=\lim_{x \rightarrow a}\frac{ e^x-e^a }{ x-a } \rightarrow f'(a)=\lim_{x \rightarrow a}\frac{ e^a(e ^{x-a}-1) }{ x-a }\] There is yet another equivalent, which comes to use here, and it looks like this: \[e ^{f(x)}-1=f(x)\] \[f(x) \rightarrow 0\] Yet again, we can observe that the exponent tends to zero, because (x-a) tends to when x-> a. So, continuing on: \[f'(a)=\lim_{x \rightarrow a}\frac{ e^a(e ^{x-a}-1) }{ x-a }\] And applying the equivalent, we can conclude: \[f'(a)=\lim_{x \rightarrow a}\frac{ e^a(e ^{x-a}-1) }{ x-a } \rightarrow f'(a)= \lim_{x \rightarrow a}\frac{ e^a (x-a) }{ (x-a) }\] \[f'(a)= \lim_{x \rightarrow a}e^a =e^a\] so, as a conclusion: \[f:f(x)=e^x \rightarrow f':f'(a)=e^a\] And we can generalize it for a point "x": \[f:f(x) = e^x \rightarrow f':f'(x)=e^x\]
yes, I'm using the very definition because it it a way of generalizing using the definition. What you did is correct, but does not include all the scenarios of the function f(x)=Ln(e^x). Try using the definition of derivative to prove more rigurously the derivative of f(x).
proofs are so nasty. How do you guys do it?
much... MUCH time of practising...

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