## Tuesday, March 14, 2006

### Derivative of e

In today's blog, I will show the proof for the derivative of e and derivative of the natural logarithm.

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

Proof:

(1) d/dx(ln x) = lim (Δx → 0) (ln (x + Δ x) - ln x)/Δ x =
= lim (Δx → 0)(ln([x + Δx]/x)/Δ x [From the basic properties of logarithms, see here]

= lim(Δx → 0) (1/Δ x)(ln([x + Δx]/x) =

= lim (Δx → 0) (ln[([x + Δx]/x)(1/Δx)])

= lim(Δx → 0) (ln[(1 + Δx/x)(1/Δx)])

(2) Let u = Δx/x

(3) lim(Δx → 0) (ln[(1 + Δx/x)(1/Δx)]) =
= lim (u → 0)(ln[(1 + u)(1/[ux])]) =
= lim(u → 0)([1/x][ln(1 + u)(1/u)])

(4) Now lim(u → 0) (1 + u)(1/u) =
= lim(u → inf)(1 + 1/u)u since:

(a) lim (u → inf)(1/u) = lim(u → 0)(u)

(b) lim (u → inf)(u) = lim (u → 0)(1/u)

(5) Now, (#4) is the definition of Euler's Number (see here), so we have:
lim(u → inf)(1 + 1/u)(u) = e

which means that:

lim(u → 0)(1 + u)(1/u) = e

(6) Putting (#5) in (#3) gives us:

lim(u → 0)([1/x][ln(1 + u)(1/u)]) =
= (1/x)ln(e) = (1/x)(1) [Since ln e = 1, see here for definition of ln ]

QED

Lemma 2: (d/dx)ex = ex

Proof:

(1) Set u = ex

(2) Using the chain rule we have:
(d/dx) ln u = (d/du)ln u * (d/dx)(ex)

(3) From lemma 1, we know that:
(d/du)ln u = (1/u)

(4) We also know that:
(d/dx)(ln(ex)) = (d/dx)(x) = 1

(5) So from #2, we get:
(d/du)ln u * (d/x)(ex) = (1/u)(d/x)(ex) = 1

(6) Or simply:
(1/u)(d/x)(ex) = 1

(7) Multiplying u to both sides gives us:
(d/x)(ex) = u

(8) But u = ex, so we get:
(d/x)(ex) = ex

QED

References:

Hi,
I require a few clarifications on 'e'. Can I mail it across to you, if I may? If so, could you please send me your mail id? Mine is jagadeeshbp@gmail.com

Anonymous said...

Your proof for d/dx (e^x) is nice, but your proof for d/dx ln(x) from first principals is invalid: the limit should be (delta x) -> 0, not x -> 0 which your proof uses (and relies on). Verify this by recognising that it is a tangent as a limit of a secant - so, as (x, f(x)) and (delta x, f(delta x)) become closer and eventually the same point.

This gives us:

d/dx (ln x) = lim (delta x -> 0) (ln(x + delta x) - ln x)/delta x
= lim ln((x + delta x) / x) / delta x
= lim (1/delta x) ln(1 + (delta x)/x)
= lim ln(1 + (delta x)/x) ^ (1/delta x)
Sub u = (delta x)/x
= lim u-> 0 ln (1 + u)^1/ux
= lim (ln(1 + u)^1/u) / x

Note that lim u->0 (1 + u)^1/u == lim n->inf (1 + 1/u)^u , which is a definition of e , so:
= ln e / x
= 1/x

Larry Freeman said...

Hi Anonymous,

Thanks very much for your comment. I looked over the proof and ouch. So many typos!

I've changed that x to Δx and also removed the 'n'.

The proof is now as it was intended.

Cheers,

-Larry