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anonymous
 one year ago
Given
x = Ln[E^x]
How do you derive the formula
d/dx E^x = E^x
with
d/dx Ln[x] = 1/x ?
anonymous
 one year ago
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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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[x=\ln(e^x)\] this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh yeah sorry, on my system all Logs are to the base E

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I know based on one rule if \[f(x) = \ln x \] then the derivative is \[f'(x) = \frac{1}{x}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1that's a theorem Theorem (The Derivative of the Natural Logarithm Function

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{d}{dx} e^x =e^x\] The derivative of \[e^x\]with respect to x is equal to e^x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1because the derivative of x is just 1 and that's normally not written down

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1an 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 \[2e^{2x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it related the the fact that E is a constant?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1this is calculus ii related. there is a bunch of log and exponential derivatives out there

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1hmmm if I use the natural log definition on this...I think something should cancel out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0perhaps I need to throw the definition of the derivative at it?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1wait to prove why e^x is the derivative of e^x. I think I typed it already

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1no no no no definition of the derivative isn't related.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I notice the derivative of 2^x is coming up as 2^x Log[2]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1that's a different rule though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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 \[\frac{ d }{ dx } E^x = E^x \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1ln e is 1? sorry I haven't done proofs in my CAlculus class....so it's a bit iffy for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah I might be wrong on that.

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.0hughfuve  your calculator only does natural logs? (Base 'e' logs?)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes sorry.. I keep forgetting that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll try to use Ln[] in future

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.0Heck, I wrote an online log calculator that will do logarithms of ANY base http://www.1728.org/logrithm.htm

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1is a proof required?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I'm still kind of new when it comes to proofs. I know some, but not all... but I know computation :)

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1So, what needs to be proven?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1derivation? at least that's on the op's question

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1so, only prove the derivative of e^x and Lnx? including Ln(e^x).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The problem reads... Start with the formulas \[x = Ln[E^x]\] and \[\frac{d}{dx} Log[x] = \frac{1}{x} \] and derive the formula \[\frac{d}{dx} e^x = e^x \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and Im assuming that means, they want some kind of proof

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1chain rule? I mean.. \[\ln(e^x) = \frac{1}{e^{x}} e^x\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1but that e^x cancels out... x.x sorry at least I tried

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah that's what I got.. I dont know what the hell they're asking for here.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1derive just means take the derivative.. the question didn't say show that or prove or disprove.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do we derive the formula, when we just know it to be true?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1If you need to prove the properties, I can show you how to prove them, but is that what the excercise requires?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry.. please remember that Log is base E

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im 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.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1So, prove the derivatives and apply it to the initial function?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1maybe 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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1taking the derivative of ln x is 1/x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1acutally 1/x (1) because the derivative of x is 1 but no one writes that these days

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[\ln x \rightarrow \frac {1}{x} (1) \rightarrow \frac{1}{x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe they want me to use logarithmic differentiation. If \[g[x] = Ln[f[x]]\] Then \[f'[x] = f[x] g'[x] \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then by the chain rule \[\frac{d}{dx} Log[f[x]] = \frac{f'[x]}{f[x]}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which is why the derivative of Log[x] = 1/x because the derivative of x = 1

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1Well... That would be an incomplete proof, the right ways is using the very definition of derivative, so, therefore: \[f:fx)=lnx\] \[f:f(x) \rightarrow f':f'(x) <=>\lim_{x \rightarrow a}\frac{ \ln(x)\ln(a) }{ xa }\] 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: \[f'(x)=\lim_{x \rightarrow a}\frac{ \ln(x)\ln(a) }{ xa } \rightarrow f'(x)=\frac{ \ln \frac{ x }{ a } }{ xa }\] 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: \[f'(x)=\lim_{x \rightarrow a}\frac{ \ln \frac{ x }{ a } }{ xa }\] We can clearly see that x/a tends to 1, because x>a. therefore, by equivalence: \[f'(x)=\lim_{x \rightarrow a}\frac{ \frac{ x }{ a }1 }{ xa }\] \[f'(x)=\lim_{x \rightarrow a}\frac{ \frac{ xa }{ a } }{ xa }\rightarrow f'(x)=\lim_{x \rightarrow a} \frac{ xa }{ a(xa) }\] \[f'(x)=\lim_{x \rightarrow a}\frac{ 1 }{ a }=\frac{ 1 }{ a }\] \[=> f'(x)=\frac{ 1 }{ a }\] And that's the derivative of any given point in the function Lnx. a generalization would be: \[f:f(x)=lnx \rightarrow f':f'(x)=\frac{ 1 }{ x }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow 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

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1The same is applied for the derivative of E^x, we will state it, then apply the definition to it: \[f:f(x)=e^x\] \[f:f(x) \rightarrow f':f'(a)<=>\lim_{x \rightarrow a}\frac{ e^xe^a }{ xa }\] 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: \[f'(a)=\lim_{x \rightarrow a}\frac{ e^xe^a }{ xa } \rightarrow f'(a)=\lim_{x \rightarrow a}\frac{ e^a(e ^{xa}1) }{ xa }\] 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 (xa) tends to when x> a. So, continuing on: \[f'(a)=\lim_{x \rightarrow a}\frac{ e^a(e ^{xa}1) }{ xa }\] And applying the equivalent, we can conclude: \[f'(a)=\lim_{x \rightarrow a}\frac{ e^a(e ^{xa}1) }{ xa } \rightarrow f'(a)= \lim_{x \rightarrow a}\frac{ e^a (xa) }{ (xa) }\] \[f'(a)= \lim_{x \rightarrow a}e^a =e^a\] so, as a conclusion: \[f:f(x)=e^x \rightarrow f':f'(a)=e^a\] And we can generalize it for a point "x": \[f:f(x) = e^x \rightarrow f':f'(x)=e^x\]

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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).

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1proofs are so nasty. How do you guys do it?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1much... MUCH time of practising...
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