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

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

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

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

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

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).

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?

sorry.. please remember that Log is base E

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

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

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

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

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

much... MUCH time of practising...