Proof: d/dx(ln x) = 1/x

[edstone61n]

This is a "sordid" video. =p

[lmq0209]

Wow, this proof is amazing!

[Nevin1343]

at 8:00, why are you able to pull the ln outside of the limit?

[flensdude]

@Nevin1343 It's called the composition law for limits, which can be summarised as:

If lim g(x) = L as x approaches a, and if the function f is continious at L, then the limit of the composite function f(g(x)) can be expressed as lim f(g(x)) = f(L) = f(lim g(x)) as x approaches a.

Example: lim ln(x + e) as x -> 0 = ln( lim(x + e) as x -> 0) = ln( 0 + e ) = ln e = 1.

[Barthayn]

@Nevin1343 it is because you are looking as what happens when u --> o. ln (1+u)--> 1

[jorelBV]

IT ALL MAKES SENSE.

[sh1tp1e]

awesomeness

[blazeeagle]

Thanks! I had problems proving derivate of ln x for myself, but everything is clear after this video. That's really fascinating.

[sbnj231t]

u r great

[dheeraj54]

is there a proof for limx->0(1+x)^(1/x)=e in one of his previous videos??

[ElessarSemaj]

There is another proof that is a lot simpler, assuming you take the approach that consists in defining e as the number so that : d(e^x)/dx = e^x, and then deducing the corresponding rules of differentiation, such as : d(e^f(x))/dx = d(f(x))/dx * e^f(x)

e^ln(x) = x (x>0) (as per definition)

d(e^ln(x))/dx = d(x)/dx

d(ln(x))/dx * e^ln(x) = 1 ( d(e^f(x))/dx = d(f(x))/dx * e^f(x) )

d(ln(x))/dx * x = 1 (e^ln(x) = x (as per definition))

d(ln(x))/dx = 1/x (since x>0)

[tIs4gatorbait]

not quite following how lim u->0 (1+u)^1/u = lim n->inf. (1+1/n)^n

[JesusHatesChristians]

@tIs4gatorbait It's simply a substitution. u in this case is defined as 1/n and inversely, n = 1/u.

n = 1/u

If u increases to infinity, then the value of n approaches zero. But if you decrease the value u (i.e., u approaches zero), then the value of n approaches infinity.

[Hatchwork]

Go do your homework or study and stop spamming.

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[rotflmaopmpqxyz]

In soviet russia, zero approaches u

[omritoker]

@rotflmaopmpqxyz Damn you're good.

Can you be my teacher? O=

[boeing747200lr]

Fascinating.

[lukaninger]

great video! helped me a lot! keep posting!

[interiorptr]

very cool! Although I watch the vids all offline I have to, once in a while, come and comment on youtube! =P Thanks Sal!

[remnant92]

hahahahaha that i awesome :P

[Marcus120489]

Thanks man helped me a lot :)

[HadesFumetas]

full color Q_Q

[MistrEgg]

I don't understand how you are able to go from lim [ln(stuff)] to ln [lim (stuff)]

[labiledm]

how you did that video-ß which software?

[Everyvideo123]

There is a much simpler way of deriving ln(x) to 1/x by simply finding the derivative with respect to y:

y=ln(x) initial equation

e^y=x making x the subject of equation

e^y=dx/dy differentiating with respect to y

x=dx/dy substituting x back into equation

we know that dy/dx=1/(dx/dy)

so dy/dx=1/x

[fpsmanager]

@Everyvideo123 Nice!

[Everyvideo123]

@fpsmanager there is only one drawback to my method, I'm assuming that the derivative of e^y is e^y, if I wanted to fully prove my method I'd have to prove fully that the derivative of e^y is e^y.

[fpsmanager]

@Everyvideo123 Which it is. If you take f(x) = e^x, integrating f(x) will give you e^x back. And it doesn't matter what variable you put into f(x), assuming y > 0 and what not of course. right?

[watch0view]

@fpsmanager the derivative of e^y is y'e^y

so you'll have to take the derivative of the exponant first and multiply that with e^y.

yeah i'm dutch really so i might've been using the wrong terms but i assume this were the right terms.

oh and indeed, if you (in dutch its called primitivate, derivitate the other way around) you get e^y back, cause you multiply by 1/y'