## Tuesday, November 1, 2016

### Unizor - Derivatives - Examples - Logarithmic Functions

Notes to a video lecture on http://www.unizor.com

Derivative Examples -
Logarithmic Functions

1. f(x) = ln(x)
(ln(x) is a natural logarithm with base e - a fundamental constant in Calculus, approximately equal to 2.71)

f I(x) = 1/x

Proof
The function increment is
ln(x+Δx)−ln(x) =
ln[(x+Δx)/x] =
ln(1+Δx/x)

Now we can use an amazing limit
(1+x)1/x → e as x→0
where e is the same fundamental constant as above.
Based on this,
ln[(1+x)1/x] → ln(e) as x→0
Using the properties of logarithms, we can transform it into
[ln(1+x)]/x → 1 as x→0
(that is, x is infinitesimal variable)

Let's use this property in calculation of our derivative.
f I(x) =
= limΔx→0[ln(1+Δx/x)]/Δx =
(substitute δ=Δx/x)
= limδ→0[ln(1+δ)]/(x·δ) =
= {limδ→0[ln(1+δ)]/δ}/x

As we noted above,
[ln(1+x)]/x → 1 as x→0
In our case the role of infinitesimal x→0 is played by variable δ.
Therefore,
limδ→0[ln(1+δ)]/δ = 1
from which we conclude
f I(x) = 1/x

2. f(x) = logb(x)

f I(x) = 1/[x·ln(b)]

Proof

We will use the following property of logarithms that allows to change the base:
logb(x) = logc(x)/logc(b)

Using this, we, firstly, convertlogb(x) into natural logarithm with base e:
logb(x) = ln(x)/ln(b)

Now we see that functionlogb(x) differs from functionln(x) only by a factor 1/ln(b).
Therefore, considering the expression for a derivative ofln(x),
f I(x) = 1/[x·ln(b)]