Dennis Bowers
James Woods
Differentiation and integration of logarithmic functions
In this paper we will go over several topics pertaining to the use of logarithmic
functions in calculus. To start with, we will prove several derivative identities involving
logarithms.
1.
(d/dx) ex= ex
Proof: 1.
y = ex
2.
ln(y) = ln(ex)
3.
(d/dx) ln(y) = y'/y = (d/dx) ln(ex) = (d/dx) x • ln(e) = (d/dx) x = 1
Line 3 uses implicit differentiation and the fact that (d/dx) f(x) = f'(x)/f(x)
4.
y' = 1•y
5.
y' = 1• ex = ex
QED
2.
(d/dx) ax = (ax) • ln(a)
Proof: 1.
y = ax
2.
ln(y) = ln(ax) = x • ln(a)
3.
(d/dx) ln(y) = y'/y = (d/dx) x • ln(a) = ln(a)
4.
y' = (ax) • ln(a)
QED
3.
(d/dx) ef(x) = f '(x)ef(x)
Proof: 1.
y = ef(x)
2.
ln(y) = ln (ef(x)) = f(x)
3.
(d/dx) y'/y = f '(x)
4.
y' = f'(x) • y = f '(x)(ef(x))
QED
4.
(d/dx) ln(x) = 1/x
Proof: 1. (d/dx) ln(x) = lim_(h  0)
derivative
2. lim_(h → 0)
(ln(x+h) – ln(x))/h by definition of
(ln(x+h) – ln(x))/h = lim_(h → 0) ln((x+h)/x)/h
3. = lim_(h → 0) 1/h ln(1+h/x)
4. let u = h/x. and xu = h
5. = lim_(u → 0) (1/xu)•ln((1+u)
6. = 1/x • lim_(u → 0) 1/u • ln(1+u)
7. = 1/x • lim_(u → 0) ln((1+u)1/u)
8. = 1/x • ln(e)
9. = 1/x
QED
While these are helpful to know, the rule that d/dx ln(f(x)) = f '(x)/f(x) is much
more helpful. This fact is used to solve many problems. Here is an example of how it's
used:
y = ln(x3cos(x)) = ln(cos(x)) + ln(x3)
y' = -sin(x)/cos(x) + 3x2/x3 =
-tan(x) + 3/x
There are reasons to use this technique even when there are no logarithms present in the
equation. One reason would be to avoid complicated applications of the product/quotient
rules such as here:
y = 2x3 • √(2x) / (1+x2) 
ln(y) = ln(2x3• √(2x) / (1+x2)) 
ln(2x3) + ln(√2x) – ln(1+x2)
ln(y)' = y'/y = (6/x + 1/(x•√2)– 2x/(1+x2))
y' = (6/x + 1/(x • √(2)) – 2x/(1+x2))•y = (6/x + 1/(x • √(2)) –
2x/(1+x2))(2x3• √(2x) / (1+x3))
Another key way it can be used is when there are variables in the exponent such as in this
situation:
y = (2x)cos(x) 
Take the natural log of both sides
ln(y) = ln((2x)cos(x)
= cos(x)ln(2x)
From there you follow the same steps as the problem before with the added use of the
product rule. Using logarithmic differentiation in this fashion seems to be its greatest
asset in practical situations, but it's also used to prove several theorems that we take for
granted such as the power rule, product rule, and quotient rule. Here are proofs to these
theorems using logarithmic differentiation:
Product rule
Quotient rule
Power rule
((d/dx) a • xn = an • x(n-1) :
y = a • xn 
ln(y) = ln(a • xn) = ln(a) + n•ln(x)
ln(y)' = y'/y = ln(a)' + n•ln(x)' = 0 + n/x
y' = n/x •y
n/x • a•xn = an•xn / x =
an • x(n-1)
QED
The other side of logarithmic differentiation is clearly logarithmic integration.
Looking at the rule ln(f(x))' = f '(x)/f(x), it's clear that the integral (f '(x)/f(x) dx) = ln|f(x)|
+ C. While that's clear, something you have to keep in mind is that the integral may not
be in exactly that form. If the exponent for every term in the numerator is 1 less than
every term in the denominator, it should clue you in that this rule applies. Here is an
example:
∫ (30x / (15x2 + 4) dx) =
ln|15x2+ 4| +C
In this example we can see how the 15x2+ 4 is the f(x) and 30x would be the f '(x). When
it's not as clear is in a situation such as this:
∫ (x / (15x2 + 4) dx)
The derivative of 15x2 + 4 doesn't = x, but we do have the proper exponents. In this case,
it's wise to look for a scalar. Since the derivative is 30x, we can write the integral in this
form and solve:
∫ (x/(15x2 + 4) dx) =
1/30 ∫ (30x / (15x2 + 4) dx) =
1/30 • ln|15x2 + 4|
The process of finding a scalar is quite simple. Take the derivative of the denominator
and compare it to the numerator. Divide the first term of the numerator by the first term
in the derivative. The result will be the potential scalar. If the numerator multiplied by
the scalar equals the derivative, then that's the scalar that will put your equation in the
form of f '(x)/f(x).
These are the basic uses of logarithmic differentiation in calculus.