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Derivatives of Logarithms
Recall that the logarithm functions satisfy very important arithmetic laws:
If a and b are positive numbers, and c is a positive number not equal to 1, we have:
logc a = ln a ln c
ln(ab) = ln a + ln b logc (ab) = logc a + logc b
ln
a b
= ln a − ln b logc
a b
= logc a − logc b
ln
ab
= b ln a logc
ab
= b logc a
The natural logarithm function was defined to the inverse of the exponential function e x^ , and the base a logarithms were defined to be the inverse of the exponential function a x^ , so we have the Cancellation Laws
eln^ x^ = x and aln^ x^ = x, so if we let y = ln x or y = loga x we have e y^ = x or a y^ = x.
Using the method of implicit differentiation, we get:
(e y^ )′^ = e y^ y′^ = (x)′^ = 1 or (ay^ )′^ = (ln a)ay^ y′^ = (x)′^ = 1,
so y′^ =
e y^
x or y′^ =
(ln a)ay^
(ln a)x
Thus
d
dx
(ln x) =
x
and
d
dx
loga x
(ln a)x
We may get more general formulas by using the Chain Rule:
d
dx
(ln f (x)) =
f ′(x)
f (x)
and
d
dx
loga f (x)
f ′(x)
(ln a)f (x)
ln a
f ′(x)
f (x)
The quantity f ′(x) f (x) is called the relative rate of change of the function f , and is
very important in practical applications.
Example 1: Find the derivative of y = ln(x^5 + x^3 + 1 ).
Solution:
y′^ = 1 x^5 + x^3 + 1 (x^5 + x^3 + 1 )′^ = 1 x^5 + x^3 + 1 ( 5 x^4 + 3 x^2 ) = x
(^2) ( 5 x (^2) + 3 ) x^5 + x^3 + 1
Example 2: Find d
dx ( sin(ln(x)).
Solution:
d dx (sin(ln(x)) = cos(ln x)
x
cos(ln x) x
Example 3: Find
d dx (ln (sin(ln(x))).
Solution:
d dx (ln (sin(ln(x))) =
1 sin (ln(x)) (sin (ln(x)))′^ =
sin(ln(x)) cos(ln(x))(ln x)′^ =
sin(ln(x)) cos(ln(x))
x
cos(ln(x)) x sin(ln(x)
Note that the domain of the derivative function is much larger than that of the original function!
Example 4: Find d
dx log 16 (sin x + cos x).
Solution:
log 16 (sin x + cos x) = ln(sin x + cos x) ln 16 , so
d dx log 16 (sin x + cos x) =
ln 16
d dx ln(sin x + cos x) =
y′ y
x
- ln π + 5 3 cos( 3 x) sin( 3 x)
x − 1
x + 2
which we need only simplify slightly to get y′^ in a usable form:
y′^ = y
[
x
- ln π + 15 cot( 3 x) − 3 x − 1
− 7.^5
x + 2
]
Example 7: Sketch the graph of y = x x
Solution:
Since, by definition, x x^ =
eln^ x^
)x = e x^ ln^ x^ , the function is only defined for x > 0, and must always be positive.
We have ln y = ln (x x^ ) = x ln x, so
y′ y = (x ln x)′^ = (x)′(ln x) + x(ln x)′^ = ( 1 ) ln x + x
x = 1 + ln x,
so y′^ = x x^ ( 1 + ln x).
The sign of y′^ depends only on that of 1 + ln x, which is 0 if ln x = −1 or x = e−^1 = 1 e
If 0 < x <
e we have y′^ < 0 and if
e < x we have y′^ > 0.
x
y
Negative x
There will be occasions when we wish to apply logarithms and deal with negative values of the variables concerned.
Of course, ln x is undefined if x ≤ 0. However, ln |x| is defined if x < 0.
Let us then find the derivative of ln |x| for non-zero x.
If x > 0, it is of course
x
If x < 0, then |x| = −x, so ln |x| = ln(−x), and we can apply the Chain Rule:
d dx (ln(−x)) =
−x
d dx (−x) =
−x
x
so we have the important formula
d
dx
(ln |x|) =
x
if x = 0
