page 81
Mathematics 0110a Summary Notes
5.5
The Differentiation of Logarithmic Functions
Let y = ln x, x > 0 ⇒ x = ey
Use implicit differentiation and the chain rule on both sides of *.
d (e y )
d ( x)
=
⇒
dx
dx
Similarly, if x < 0,
So
d ln x
dx
1 = ey
dy
dy
1
⇒
= y ⇒
e
dx
dx
dy d ln x 1
=
= .
x
dx
dx
d ln(− x) (−1) 1
=
= using the chain rule.
dx
−x
x
=
1
x
Generalizing, using the chain rule with u = u(x) gives
d (ln u )
1 du
=
dx
u dx
Examples:
Find
dy
for each of the following:
dx
[( x 2 + 1)( x − 1) 4 ]
x3 ( x + 1)
1. y = ln(x2 + 4)
2. y = ln ex
3. y = ln
4. y = ex (ln x)5
5. y = x2 ln (x5)
6. y = ln |x|
⎛1⎞ 1
7. y = ln ⎜ ⎟ +
⎝ x ⎠ ln x
8. x ln y = x2
9. ln (xy) = ex
Example 10. Find the equation of the tangent line to the curve y = ln(ln x) which is
perpendicular to ex + y = 0.
solution:
We want y′ =
1 ⎛1⎞ 1
⎜ ⎟ = or x ln x = e so x = e.
ln x ⎝ x ⎠
e
y(e) = ln (ln e) = ln 1 = 0
The required equation is: y − 0 =
1
x
(x − e) or y = − 1.
e
e
page 82
Mathematics 0110a Summary Notes
•
General Exponential Functions
Consider once again, y = bx, where b > 0.
x
Change the base to e, y = e ln b = exln b.
So y ′ = exln b (ln b) = bx ln b.
d (b x )
= bx ln b
dx
Generalizing, using the chain rule with u = u(x) gives
d (bu )
du
= bu ln b
dx
dx
Example 11.
d 5x
= 5 x ln 5
dx
Example 12.
d 10ln x
ln10
= 10ln x
dx
x
•
General Logarithmic Functions
Consider now, y = log b x , where b > 0.
Then x = by
by definition.
Now use implicit differentiation.
1 = by ln b y ′
So y ′ =
1
1
=
.
b ln b
x ln b
y
Mathematics 0110a Summary Notes
page 83
Then
d (log b x)
1
=
dx
x ln b
.
Generalizing, using the chain rule with u = u(x) gives
d (log b u )
1 du
=
dx
u ln b dx
Example 13.
d log 5 x
1
=
dx
x ln 5
d log 2 ( x 2 + e x )
2x + ex
Example 14.
= 2 x
dx
( x + e ) ln 2
Another approach to developing the formulas above is simply to use the change of base
formula.
Recall, log b x =
d ⎛ ln x ⎞
d (log b x)
ln x
1
so
=
.
⎜
⎟=
dx ⎝ ln b ⎠ x ln b
dx
ln b
With base b ≠ e, derivatives of general logs and exponentials proceed as before with the
added term ln b. Obviously when b = e, ln b = ln e = 1 indicating once again why e is a
natural base.
Example 15. Given f(x) = log 4 (log 2 x) find f ′(16).
solution:
f ′(x) =
d (log 4 (log 2 x)
1
1
1
=
dx
log 2 x x ln 2 ln 4
f ′(16) =
1 1
1
1
1
=
4 16 ln 2 ln 4 64 ln 2 ln 4
page 84
Mathematics 0110a Summary Notes
•
Logarithmic Differentiation
The properties of logarithms make them useful tools for the differentiation of
complicated functions that consist of products, quotients and exponential or
combinations of these.
•
procedure:
1. determine whether log differentiation applies
− expression is a combination of products, quotients, exponentials
− sums and differences of terms are not eligible
− y written explicitly in terms of x
2.
take the natural log of both sides
3.
simplify the rhs as much as possible using log properties
4.
differentiate term by term (implicitly on the lhs)
dy
5.
solve for
dx
Example 16: Let y =
3x ( x 2 − 2)3 ( x + 3)
(1 − x) x + 2 x
3
.
Find
dy
.
dx
solution: 1. y is expressed solely in terms of products, quotients and
exponentials.
2. ln y = ln
3x ( x 2 − 2)3 ( x + 3)
(1 − x) x3 + 2 x
3. = x ln 3 + 3ln (x2 − 2) + ln (x + 3) − ln (1 − x) − (1/2) ln (x3 + 2x)
4.
5.
1 dy
2x
1
(−1) 1 3x 2 + 2
= ln 3 + 3 2
+
−
−
y dx
x − 2 x + 3 1 − x 2 x3 + 2 x
⎛
dy
2x
1
(−1) 1 3x 2 + 2 ⎞
= y ⎜ ln 3 + 3 2
+
−
−
⎟
dx
x − 2 x + 3 1 − x 2 x3 + 2 x ⎠
⎝
3x ( x 2 − 2)3 ( x + 3)
=
(1 − x) x3 + 2 x
⎛
6x
1
1
3x 2 + 2 ⎞
+
+
+
−
ln
3
⎜
⎟
x 2 − 2 x + 3 1 − x 2( x3 + 2 x) ⎠
⎝
page 85
Mathematics 0110a Summary Notes
In the last example, the application of ln to both sides converts products and quotients
into sums and differences − simplifying the differentiation.
Note that
•
d
1 dy
ln y =
on the lhs.
dx
y dx
generalized exponentials
The properties of logarithms also simplifies the differentiation of exponential
expressions in which both the base and exponent are non−constant.
Taking logarithms converts these expressions into simple products. Differentiation
can then proceed using the product rule.
Example 2:
solution:
Let y = x x + ln x .
Find
dy
.
dx
1. The expression is an exponential function.
2. ln y = ln x x + ln x
3. ln y = (x + ln x) ln x
•
4.
1 dy
1
1
= (1 + ) ln x + (x + ln x)
y dx
x
x
5.
dy
1
1
2
= y [(1 + ) ln x + (x + ln x) ] = x x + ln x [ ln x + ln x + 1]
x
x
x
dx
the power rule for real values of the exponent
Let y = xr, where r is any real number.
Use log differentiation.
ln y = ln xr = r ln x
1 dy
r
=
y dx
x
xr
dy
r
= y
= r
= r x r −1
dx
x
x
Therefore:
d r
du
u = ru r −1
dx
dx
as expected.
by applying the chain rule.