Differentiating logarithm functions.
To differentiate y = ln x, we must return to the definition...
ln(x + h) − ln x
d
(ln x) = lim
h→0
dx
h
and simplify, using algebraic properties of ln x...
1
= lim (ln(x + h) − ln x)
h→0 h
1
x+h
= lim ln
h→0 h
x
"
1/h #
h
= lim ln 1 +
h→0
x
and use the continuity of ln x...
"
1/h #
h
= ln lim 1 +
h→0
x
What next?
1
Remember the special limit
lim (1 + u)1/u = e . . .
u→0
and substitute (rename)
h
1
1
1 1
= u =⇒ h = ux =⇒
=
= · ...
x
h
ux
u x
So
lim
h→0
h
1+
x
1/h
= lim (1 +
u→0
1 1
·
u) u x
= lim
u→0
(1 +
=
lim (1 +
u→0
1
u) u
1
u) u
1
x
1
x
(because f (x) = a1/x is continuous)
= e1/x
2
Returning to
d
(ln x)...
dx
d
ln(x + h) − ln x
(ln x) = lim
h→0
dx
h
..
.
"
1/h #
h
= ln lim 1 +
h→0
x
= ln(e1/x )
1
= .
x
I.e.,
d
1
(ln x) =
dx
x
3
Example 1.
1
d
2
3x ln x = 6x ln x + 3x2 · = 6x ln x + 3x
dx
x
(product rule)
Example 2.
d
1
10x + 3
2
ln 5x + 3x + 1 = 2
· (10x + 3) = 2
dx
5x + 3x + 1
5x + 3x + 1
(chain rule)
2
Example 3. Differentiate y = ln 5x .
We can use the chain rule again:
1
10x
2
y = 2 · 10x = 2 = . . .
5x
5x
x
0
or we can simplify and then differentiate:
y = ln 5x
2
1
2
= ln 5 + ln x = ln 5 + 2 ln x =⇒ y = 0 + 2 · =
x
x
0
2
4
Example 4. Find the equation of the tangent line to the graph
2
x +x−1
y = ln
4x − 3
at the point where x = 1.
1+1−1
Point: y(1) = ln
= ln 1 = 0, so the point is (1, 0).
4−3
Slope: We need to compute y 0 (1), and once again, it is easier to simplify
before we differentiate:
2
x +x−1
y = ln
= ln(x2 + x − 1) − ln(4x − 3)
4x − 3
2x + 1
4
=⇒ y = 2
−
x + x − 1 4x − 3
3 4
=⇒ y 0 (1) = − = −1.
1 1
So the equation of the tangent line is
0
y = y(1) + y 0 (1)(x − 1) =⇒ y = −(x − 1) =⇒ y = 1 − x.
5
Figure 1: Graph of y = ln
x2 +x−1
4x−3
6
and its tangent line at (1, 0)
Logarithmic differentiation:
If f (x) is any (differentiable) function, then
1
f 0 (x)
d
0
ln(f (x)) =
· f (x) =
.
dx
f (x)
f (x)
This is called the logarithmic derivative of f (x).
d
ln(f (x)) than it is to
In some cases, it is easier to compute the dx
compute f 0 (x), because ln(f (x)) is a simpler function than f (x) itself.
We can take advantage of this to find f 0 (x) because
d
f 0 (x) = f (x) ·
ln(f (x)) .
dx
This is called logarithmic differentiation.
We will use this idea in the following special, but important case...
7
Differentiating the exponential function.
Observation: ln(ex ) = x which is a simpler function (to differentiate)
d x
x
e :
than e . So we will use logarithmic differentiation to find
dx
d x
d
d
e = ex ·
(ln(ex )) = ex ·
x = ex · 1 = ex .
dx
dx
dx
I.e.,
d x
e = ex
dx
Example 5.
3 x −1/2
d x 1/2 x 1/2
x 1/2
x 1 −1/2
= 3e x + e x
3e x
= 3e x + 3e · x
dx
2
2
product rule.
Example 6.
ex
2
−3x+1
0
= ex
chain rule
8
2
−3x+1
(2x − 3)
Example 7. Find the marginal revenue function for the firm whose
demand equation is given by
p = 15e2−0.05q .
First, find the revenue function
r = pq = 15qe2−0.05q .
Now differentiate, using the product and chain rules:
dr
= 15e2−0.05q + 15qe2−0.05q (−0.05) = (15 − 0.75q)e2−0.05q .
dq
9
The consumption function for a small country is given by
0.95Y e
C = ln 0.2Y
,
e
+5
where Y is national income, measured in $billions.
(a) How much is consumed when Y = 10?
9.5 e
C(10) = ln 2
≈ 6.983
e +5
(b) What is the marginal propensity to consume when Y = 10?
0.95Y dC
d
e
=
ln 0.2Y
dY
dY
e
+5
d
0.95Y
0.2Y
=
ln e
− ln e
+5
dY
0.2Y
0.2e
d
=
0.95Y − ln e0.2Y + 5 = 0.95 − 0.2Y
dY
e
+5
dC 0.2e2
= 0.95 − 2
≈ 0.8307
dY Y =10
e +5
10
dC
, and interpret the result.
Y →∞ dY
dC
0.2e0.2Y
lim
= lim 0.95 − 0.2Y
Y →∞ dY
Y →∞
e
+5
(c) Compute the limit
lim
0.2e0.2Y · e−0.2Y
= 0.95 − lim
Y →∞ (e0.2Y + 5) · e−0.2Y
0.2
0.2
= 0.75
= 0.95 − lim
=
0.95
−
Y →∞ 1 + 5e−0.2Y
1
Interpretation: When income grows large, the nation will tend to
consume $0.75 of each additional dollar of income.
11