Exponential Diffrentiation (1 Viewer)

[jaychouf4n]

Simplify then diffrentiate e^log x

ok so i'm not sure what they mean by simplify

i get e^log x/x by chain rule

the answer in the back is 1

Can you please explain the reasoning for me

Here log is ln because it is cambridge

 

[Iruka]

The logarithmic and exponential functions are inverses of each other, so e^logx = x.

 

[Mark576]

The exponential and natural logarithmic functions are mutually inverse, so they will cancel each other when applied together. So e^{ln x} = x, and the derivative is then obviously 1, as required.

 

[Pwnage101]

uh, isn't it pretty obvious....

e^(lnx) = x, by definition

 

[jaychouf4n]

omg!!

I'm such an idiot

-hits self in head-

 

[Forbidden.]

uh, isn't it pretty obvious....

e^(lnx) = x, by definition

Prove it.

 

[lolokay]

Prove it.

e to the power [the power e must be put to to make it equal to x] = x

or you could say
ln x = a
e^a = x
e^ln x = e^a = x

but the definition is sufficient imo

 

[lolokay]

eh? ln x = a, therefore e^a = x. is true by a definition. I was just saying that e^ln x = x is also true basically by definition, so you don't really need to prove it

 

[Aerath]

Can't you do?
RTP: e^lnx = x
Let y = e^lnx
(Log both sides)
ln y = ln e^lnx
lny = lnx*lne
lny = ln x
y = x

Therefore:
e^lnx = x

Or am I totally off track? =\

 

[vds700]

Can't you do?
RTP: e^lnx = x
Let y = e^lnx
(Log both sides)
ln y = ln e^lnx
lny = lnx*lne
lny = ln x
y = x

Therefore:
e^lnx = x

Or am I totally off track? =\

thats the way i would do it, its fine.