Differentiating ln x from first principles

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  • #1
Imperial

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Hi all.

For any of you who have done differential calculus, I need a little help with a problem involving natural logarithms.

The question asks to differentiate y = ln x from first principles . It says "use the definition of the Euler number, namely e = lim(n->inf.) (1+1/n)^n.".
First principles means f'(x) = lim(h->0) [f(x+h) - f(x)] / h (this is the first thing we learned in calculus).

I so far managed two different methods:
Method 1. y = ln x
therefore e^y = x
dx/dy = e^y.
Since dx/dy * dy/dx = 1
1/(dx/dy) = dy/dx.
= 1/e^y
= 1/x.

Method 2. y = ln x
f(x) = ln x
f(x+h) = ln (x+h)
f'(x) = lim(h->0) [ln (x+h) - ln x] / h
= lim(h->0) ln (x+h/x) / h
= lim(h->0) 1/h * ln(1+h/x)
Since lim(h->0) ln(1+h/x) -> h/x where h != 0,
f'(x) = 1/h * h/x
= 1/x

Both of these methods work and are valid, although I didn't bring the definition of the Euler number into it. I personally have no idea how to do this. Could anyone here who has done a bit of math before please help me with this?

Thanks.
 

Answers and Replies

  • #2
HallsofIvy
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Would you mind telling me how you arrive at
= lim(h->0) ln (x+h/x) / h
= lim(h->0) 1/h * ln(1+h/x)
?
What happened to the first x?

Not to mention:
Since lim(h->0) ln(1+h/x) -> h/x where h != 0
Surely you know that lim(h->0) does NOT depend on h!


Probably what your instructor intends is to use your first calculation: if y= ln(x) then x= e^y . HOWEVER, you cannot assume that the derivative of e^x is e^x: that's where "first principles" comes in.

If y(x)= e^x, then y(x+ h)= e^(x+h) so (y(x+h)- y(x))/h=
(e^(x+h)-e^x)/h= (e^x e^h- e^x)/h= (e^x)((e^h-1)/h).

The derivative of e^x is (lim(h->0)(e^h-1)/h) e^x.

You need to show "from first principles" that

lim(h->0) (e^h-1)/h = 1.
 
  • #3
Imperial
The first "x" wasn't lost- remember the logarithm laws:
ln a - ln b = ln(a/b) so my expression ln(x+h) - ln(x) became ln(x+h/x).

As for differentiating e^x, that is easy.
f(x) = e^x
f(x+h) = e^(x+h)
f'(x) = lim(h->0) e^(x+h) - e^(x) / h
= lim(h->0) e^x(e^h - 1) / h
= e^x lim(h->0) e^h - 1 /h
The definition of the Euler number is lim(n->inf.) (1+1/n)^n which can become lim(h->0) (1+h)^1/h.
therefore
f'(x) = e^x lim(h->0) (1+h)^1/h^h - 1 / h
= e^x lim(h->0) 1 + h - 1 /h
= e^x lim(h->0)h/h
= e^x

Although that doesn't really come into it. A previous question in the excercise asked that, and I think that there is probably some other method which needs to be used. Is anyone here familiar with any other methods?
 
  • #4
HallsofIvy
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Oh, I see now: when you wrote ln(x+h/x) you MEANT ln((x+h)/x)
(although when you wrote ln(1+ h/x) you DIDN'T mean ln((1+h)/x).)
 

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