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)e^x = e^x

Proof:

(1) Set u = e^x

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

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

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

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

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

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

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

QED

References:

• (d/dx)ln x, math.com
• (d/dx)e^x,math2.org