18. DERIVATIVES OF LOGARITHMIC FUNCTIONS
55
17.3. Exercises.
(1) Let C be the circle given by the equation x2 + y 2 = 169. Use
implicit differentiation to find the slope of the tangent line to
C at the point (5, 12).
1
(2) Prove that (sin−1 x) = √1−x
2.
1
−1
√
(3) Prove that (cos x) = − 1−x2 .
1
(4) Prove that (cot−1 x) = − 1+x
2.
1
−1
(5) Prove that (sec x) = x√x2 −1 .
(6) Prove that (csc−1 x) = − x√x12 −1 .
18. Derivatives of Logarithmic Functions
18.1. The Formula for (loga x) . As another powerful application of im-
plicit differentiation, we compute the derivative of the function f (x) =
ln x.
Theorem 21. We have
1
.
x
Proof. Set y = ln x. Then ey = x. Differentiating both sides with
respect to x, we get
(ln x) =
ey ·
dy
= 1,
dx
1
dy
= y.
dx
e
However, ey = x by definition, so
dy
1
=
dx
x
as claimed.
2
It is now a breeze to determine the derivative of logarithmic functions of any base.
Corollary 3. Let a = 1 be a fixed positive real number. Then
1
.
(loga x) =
x ln a
Proof. Note that
log x
x = eln a a = e(ln a)(loga x) .
56
3. RULES OF DIFFERENTIATION
So ln x = (ln a) · (loga x) and
f (x) = loga x =
ln x
.
ln a
As ln a is a constant, it follows that
1
1
f (x) =
(ln x) =
ln a
x ln a
as claimed.
2
18.2. The Chain Rule and ln x. An interesting consequence of Theorem
21 is the following.
Corollary 4. Let f (x) be a differentiable function that takes positive values only. Then
f (x)
d
ln f (x) =
.
dx
f (x)
Proof. By the chain rule,
d df
d
f (x)
ln f =
.
·
=
dx
df dx
f (x)
Example 32. Let f (x) = cos x. Then
2
d
− sin x
= − tan x.
ln(cos x) =
dx
cos x
18.3. Logarithmic Differentiation. Sometimes we need to compute the
derivative of a complicated product. This is sometimes easier by taking
the logarithm of the product, which will be a sum, and using implicit
differentiation. This procedure, which is called logarithmic differentiation, has the inherent advantage that it deals with sums instead of
products, and sums are much easier to differentiate than products.
Example 33. Let
√
x3 x + 1
.
y= √
x−2
Compute dy/dx.
Solution: Taking logarithms, we get
1
1
ln y = 3 ln x + ln(x + 1) − ln(x − 2).
2
2
Now taking derivatives with respect to x and using Corollary 4, we
have
dy 1
3
1
1
· = +
−
.
dx y
x 2(x + 1) 2(x − 2)
18. DERIVATIVES OF LOGARITHMIC FUNCTIONS
Finally, we can solve this equation for dy/dx to get
3
1
1
dy
=y
+
−
dx
x 2(x + 1) 2(x − 2)
√
3
x3 x + 1
1
1
.
·
= √
+
−
x 2(x + 1) 2(x − 2)
x−2
57
2
18.4. Power Functions Revisited. Recall that in an earlier section, we
proved that if n is a fixed positive integer, then (xn ) = nxn−1 . We
stated that this was the case for all real numbers n, not just positive
integers, but we have not proved that claim. Now we have the tools,
namely logarithmic differentiation, to prove it.
Theorem 22. Let n be any real number. Then we have
d n
x = nxn−1 .
dx
Proof. Set y = xn . Let us assume for the case of simplicity that
x is positive. Taking logarithms, we have
ln y = n ln x.
Differentiating both sides with respect to x, we get
n
dy 1
· = .
dx y
x
Solving for dy/dx yields
dy
ny
nxn
=
=
= nxn−1
dx
x
x
as claimed.
2
18.5. The Number e Revisited. Recall that we have defined the number e, the base of the natural logarithm, as the number for which
limh→0 (eh − 1)/h = 1. Our new knowledge lets us express e more directly, as a limit.
Note that if f (x) = ln x, then f (x) = 1/x, so f (1) = 1. By the
definition of derivatives, this means that
ln(1 + h) − ln 1
= 1.
lim
h→0
h
Observing that ln 1 = 0 and using the power rule of logarithms, we get
lim ln(1 + h)1/h = 1,
h→0
or, applying the exponential function ez to both sides, we have
lim (1 + h)1/h = e.
h→0
58
3. RULES OF DIFFERENTIATION
Equivalently, setting x = 1/h, we get
x
1
= e.
lim 1 +
x→∞
x
Either of the last two formulas can help to determine the approximate
value 2.712828 of e.
18.6. Exercises.
√
d
Compute dx
ln( x + 1).
Compute (ln |x|) .
Compute (xx ) .
√
3
x+4
.
Compute f (x) if f (x) = x4 x+1
1 2x
(5) Compute limx→∞ 1 + x .
(1)
(2)
(3)
(4)
19. Applications of Rates of Change
In this section, we consider a few applications of derivatives in various disciplines.
19.1. Physics. Recall that if an object moves along a line and the
distance it covers in time t is described by the function s(t), then
s(t + h) − s(t)
ds
= s (t) = lim
h→0
dt
h
is the instantaneous velocity of the object at time t.
We can take this concept one step further. If the object moves at
a changing velocity, then the rate of change of the velocity itself can
be important information. For instance, when considering a vehicle’s
performance, we may be interested in how fast it can reach its top
speed, not only what its top speed is.
The corresponding notion in physics is called acceleration, and is
denoted by a(t). That is, keeping the previous notation, we have
(3.14)
v(t) =
dv
= s (t).
dt
Example 34. The position of a particle is described by the equation
1
(3.16)
s(t) = t3 − 3t2 + 5t.
3
Here s is measured in meters and t in seconds.
(I) What is the velocity of the particle after 3 seconds?
(II) Find the acceleration of the particle after 10 seconds.
(III) When does the particle move backward?
(3.15)
a(t) = v (t) =