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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 ?

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  1. UsukiDoll
    • one year ago
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    \[x=\ln(e^x)\] this?

  2. anonymous
    • one year ago
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    oh yeah sorry, on my system all Logs are to the base E

  3. UsukiDoll
    • one year ago
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    I know based on one rule if \[f(x) = \ln x \] then the derivative is \[f'(x) = \frac{1}{x}\]

  4. UsukiDoll
    • one year ago
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    that's a theorem Theorem (The Derivative of the Natural Logarithm Function

  5. UsukiDoll
    • one year ago
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    \[\frac{d}{dx} e^x =e^x\] The derivative of \[e^x\]with respect to x is equal to e^x

  6. UsukiDoll
    • one year ago
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    because the derivative of x is just 1 and that's normally not written down

  7. UsukiDoll
    • one year ago
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    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 \[2e^{2x}\]

  8. anonymous
    • one year ago
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    is it related the the fact that E is a constant?

  9. UsukiDoll
    • one year ago
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    errr...not exactly

  10. UsukiDoll
    • one year ago
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    this is calculus ii related. there is a bunch of log and exponential derivatives out there

  11. UsukiDoll
    • one year ago
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    hmmm if I use the natural log definition on this...I think something should cancel out

  12. anonymous
    • one year ago
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    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

  13. anonymous
    • one year ago
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    perhaps I need to throw the definition of the derivative at it?

  14. UsukiDoll
    • one year ago
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    wait to prove why e^x is the derivative of e^x. I think I typed it already

  15. UsukiDoll
    • one year ago
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    no no no no definition of the derivative isn't related.

  16. anonymous
    • one year ago
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    I notice the derivative of 2^x is coming up as 2^x Log[2]

  17. anonymous
    • one year ago
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    er Ln[2]

  18. UsukiDoll
    • one year ago
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    that's a different rule though

  19. anonymous
    • one year ago
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    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 \[\frac{ d }{ dx } E^x = E^x \]

  20. UsukiDoll
    • one year ago
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    ln e is 1? sorry I haven't done proofs in my CAlculus class....so it's a bit iffy for me

  21. anonymous
    • one year ago
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    yeah I might be wrong on that.

  22. anonymous
    • one year ago
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    x = Log[E^x]

  23. anonymous
    • one year ago
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    Ln[E] = 1 is true

  24. wolf1728
    • one year ago
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    hughfuve - your calculator only does natural logs? (Base 'e' logs?)

  25. anonymous
    • one year ago
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    yes sorry.. I keep forgetting that.

  26. anonymous
    • one year ago
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    I'll try to use Ln[] in future

  27. wolf1728
    • one year ago
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    Heck, I wrote an online log calculator that will do logarithms of ANY base http://www.1728.org/logrithm.htm

  28. anonymous
    • one year ago
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    cool stuff

  29. wolf1728
    • one year ago
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    gee thanks :-)

  30. Owlcoffee
    • one year ago
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    is a proof required?

  31. UsukiDoll
    • one year ago
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    I'm still kind of new when it comes to proofs. I know some, but not all... but I know computation :)

  32. Owlcoffee
    • one year ago
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    So, what needs to be proven?

  33. UsukiDoll
    • one year ago
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    derivation? at least that's on the op's question

  34. Owlcoffee
    • one year ago
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    so, only prove the derivative of e^x and Lnx? including Ln(e^x).

  35. anonymous
    • one year ago
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    The problem reads... Start with the formulas \[x = Ln[E^x]\] and \[\frac{d}{dx} Log[x] = \frac{1}{x} \] and derive the formula \[\frac{d}{dx} e^x = e^x \]

  36. anonymous
    • one year ago
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    and Im assuming that means, they want some kind of proof

  37. UsukiDoll
    • one year ago
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    chain rule? I mean.. \[\ln(e^x) = \frac{1}{e^{x}} e^x\]

  38. UsukiDoll
    • one year ago
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    but that e^x cancels out... x.x sorry at least I tried

  39. anonymous
    • one year ago
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    yeah that's what I got.. I dont know what the hell they're asking for here.

  40. UsukiDoll
    • one year ago
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    derive just means take the derivative.. the question didn't say show that or prove or disprove.

  41. Owlcoffee
    • one year ago
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    I'm confused.

  42. anonymous
    • one year ago
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    how do we derive the formula, when we just know it to be true?

  43. Owlcoffee
    • one year ago
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    If you need to prove the properties, I can show you how to prove them, but is that what the excercise requires?

  44. anonymous
    • one year ago
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    sorry.. please remember that Log is base E

  45. anonymous
    • one year ago
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    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.

  46. Owlcoffee
    • one year ago
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    So, prove the derivatives and apply it to the initial function?

  47. UsukiDoll
    • one year ago
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    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

  48. UsukiDoll
    • one year ago
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    taking the derivative of ln x is 1/x

  49. UsukiDoll
    • one year ago
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    acutally 1/x (1) because the derivative of x is 1 but no one writes that these days

  50. UsukiDoll
    • one year ago
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    \[\ln x \rightarrow \frac {1}{x} (1) \rightarrow \frac{1}{x}\]

  51. anonymous
    • one year ago
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    Maybe they want me to use logarithmic differentiation. If \[g[x] = Ln[f[x]]\] Then \[f'[x] = f[x] g'[x] \]

  52. anonymous
    • one year ago
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    then by the chain rule \[\frac{d}{dx} Log[f[x]] = \frac{f'[x]}{f[x]}\]

  53. anonymous
    • one year ago
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    which is why the derivative of Log[x] = 1/x because the derivative of x = 1

  54. Owlcoffee
    • one year ago
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    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: \[f:f(x)=lnx \rightarrow f':f'(x)=\frac{ 1 }{ x }\]

  55. anonymous
    • one year ago
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    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

  56. Owlcoffee
    • one year ago
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    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": \[f:f(x) = e^x \rightarrow f':f'(x)=e^x\]

  57. Owlcoffee
    • one year ago
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    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).

  58. UsukiDoll
    • one year ago
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    proofs are so nasty. How do you guys do it?

  59. Owlcoffee
    • one year ago
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    much... MUCH time of practising...

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