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

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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- katieb

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- UsukiDoll

\[x=\ln(e^x)\] this?

- anonymous

oh yeah sorry, on my system all Logs are to the base E

- UsukiDoll

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

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## More answers

- UsukiDoll

that's a theorem
Theorem (The Derivative of the Natural Logarithm Function

- UsukiDoll

\[\frac{d}{dx} e^x =e^x\]
The derivative of \[e^x\]with respect to x
is equal to e^x

- UsukiDoll

because the derivative of x is just 1 and that's normally not written down

- UsukiDoll

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
\[2e^{2x}\]

- anonymous

is it related the the fact that E is a constant?

- UsukiDoll

errr...not exactly

- UsukiDoll

this is calculus ii related. there is a bunch of log and exponential derivatives out there

- UsukiDoll

hmmm if I use the natural log definition on this...I think something should cancel out

- anonymous

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

- anonymous

perhaps I need to throw the definition of the derivative at it?

- UsukiDoll

wait to prove why e^x is the derivative of e^x. I think I typed it already

- UsukiDoll

no no no no definition of the derivative isn't related.

- anonymous

I notice the derivative of 2^x is coming up as 2^x Log[2]

- anonymous

er Ln[2]

- UsukiDoll

that's a different rule though

- anonymous

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
\[\frac{ d }{ dx } E^x = E^x \]

- UsukiDoll

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

- anonymous

yeah I might be wrong on that.

- anonymous

x = Log[E^x]

- anonymous

Ln[E] = 1 is true

- wolf1728

hughfuve - your calculator only does natural logs? (Base 'e' logs?)

- anonymous

yes sorry.. I keep forgetting that.

- anonymous

I'll try to use Ln[] in future

- wolf1728

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

- anonymous

cool stuff

- wolf1728

gee thanks :-)

- Owlcoffee

is a proof required?

- UsukiDoll

I'm still kind of new when it comes to proofs. I know some, but not all... but I know computation :)

- Owlcoffee

So, what needs to be proven?

- UsukiDoll

derivation? at least that's on the op's question

- Owlcoffee

so, only prove the derivative of e^x and Lnx?
including Ln(e^x).

- anonymous

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

- anonymous

and Im assuming that means, they want some kind of proof

- UsukiDoll

chain rule?
I mean..
\[\ln(e^x) = \frac{1}{e^{x}} e^x\]

- UsukiDoll

but that e^x cancels out... x.x sorry at least I tried

- anonymous

yeah that's what I got.. I dont know what the hell they're asking for here.

- UsukiDoll

derive just means take the derivative.. the question didn't say show that or prove or disprove.

- Owlcoffee

I'm confused.

- anonymous

how do we derive the formula, when we just know it to be true?

- Owlcoffee

If you need to prove the properties, I can show you how to prove them, but is that what the excercise requires?

- anonymous

sorry.. please remember that Log is base E

- anonymous

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.

- Owlcoffee

So, prove the derivatives and apply it to the initial function?

- UsukiDoll

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

- UsukiDoll

taking the derivative of ln x is 1/x

- UsukiDoll

acutally
1/x (1)
because the derivative of x is 1 but no one writes that these days

- UsukiDoll

\[\ln x \rightarrow \frac {1}{x} (1) \rightarrow \frac{1}{x}\]

- anonymous

Maybe they want me to use logarithmic differentiation.
If
\[g[x] = Ln[f[x]]\]
Then
\[f'[x] = f[x] g'[x] \]

- anonymous

then by the chain rule
\[\frac{d}{dx} Log[f[x]] = \frac{f'[x]}{f[x]}\]

- anonymous

which is why the derivative of
Log[x] = 1/x
because the derivative of x = 1

- Owlcoffee

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:
\[f:f(x)=lnx \rightarrow f':f'(x)=\frac{ 1 }{ x }\]

- anonymous

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

- Owlcoffee

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":
\[f:f(x) = e^x \rightarrow f':f'(x)=e^x\]

- Owlcoffee

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).

- UsukiDoll

proofs are so nasty. How do you guys do it?

- Owlcoffee

much... MUCH time of practising...

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