Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

\[2e^{2x}\]

Answer 8

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

Answer 9

errr...not exactly

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

er Ln[2]

Answer 18

that's a different rule though

Answer 19

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

Answer 20

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

Answer 21

yeah I might be wrong on that.

Answer 22

x = Log[E^x]

Answer 23

Ln[E] = 1 is true

Answer 24

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

Answer 25

yes sorry.. I keep forgetting that.

Answer 26

I'll try to use Ln[] in future

Answer 27

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

Answer 28

cool stuff

Answer 29

gee thanks :-)

Answer 30

is a proof required?

Answer 31

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

Answer 32

So, what needs to be proven?

Answer 33

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

Answer 34

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

Answer 35

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

and derive the formula
\[\frac{d}{dx} e^x = e^x \]

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

I'm confused.

Answer 42

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

Answer 43

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

Answer 44

sorry.. please remember that Log is base E

Answer 45

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.

Answer 46

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

Answer 47

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

Answer 48

taking the derivative of ln x is 1/x

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

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).

Answer 58

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

Answer 59

much... MUCH time of practising...