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Unizor - Derivatives - Examples - Logarithmic Functions
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Unizor - Creative Minds through Art of Mathematics - Math4Teens
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'(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'(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'(x) = 1/x
2. f(x) = log_b(x)
f'(x) = 1/[x·ln(b)]
Proof
We will use the following property of logarithms that allows to change the base:
log_b(x) = log_c(x)/log_c(b)
Using this, we, firstly, convert log_b(x) into natural logarithm with base e:
log_b(x) = ln(x)/ln(b)
Now we see that function log_b(x) differs from function ln(x) only by a factor 1/ln(b).
Therefore, considering the expression for a derivative of ln(x),
f'(x) = 1/[x·ln(b)]
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'(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'(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'(x) = 1/x
2. f(x) = log_b(x)
f'(x) = 1/[x·ln(b)]
Proof
We will use the following property of logarithms that allows to change the base:
log_b(x) = log_c(x)/log_c(b)
Using this, we, firstly, convert log_b(x) into natural logarithm with base e:
log_b(x) = ln(x)/ln(b)
Now we see that function log_b(x) differs from function ln(x) only by a factor 1/ln(b).
Therefore, considering the expression for a derivative of ln(x),
f'(x) = 1/[x·ln(b)]
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