1. Logarithmic and exponential functions
In this section, we review some basic properties of the logarithmic and exponential functions.
Logarithms (for a > 0)
(1) y = loga (x) ↔ x = ay
(2) loga (xy) = loga (x) + loga (y)
(3) loga ( xy ) = loga (x) − loga (y)
(4) loga (xp ) = p loga (x)
(5) loga
√
p
x=
loga (x)
p
(6) loga (a) = 1
(7) loga (1) = a
(8) loga (ax ) = x
Logarithms with base e
Recall that when our base is Euler’s constant e, we denote it loge (x) = ln(x) and
is called the natural logarithm. The natural logarithm satisfies all the properties
above as well. Notice that y = ln(x) ↔ x = ey and ln(e) = 1.
Exponential functions (for a > 0)
(1) y = ax ↔ x = loga (y)
(2) ax ay = ax+y
(3)
ax
ay
= ax−y
(4) (ax )y = axy
x
(5) (ax )1/y = a y
(6) a0 = 1
1
2
(7) a1 = a.
(8) aloga (x) = x
Exponential functions with base e
Again, for base e, ex satisfies all the same properties above.
2. Derivatives for logarithms and exponentials
Theorem 2.1 (Derivatives for logarithms). Suppose f (x) = loga (x). Then f ′ (x) =
1
1
. For base e, if f (x) = ln(x), then f ′ (x) = .
(ln a)x
x
Example 1. Let f (x) = ln(x + 4). Find f ′ (x).
Notice that we have a compostition of functions. The outside function is g(x) =
ln(x), and the inside function is h(x) = x + 4. By the chain rule, we have f ′ (x) =
1
1
⋅1=
.
x+4
x+4
Example 2. Let f (x) = log5 (x2 ). Find f ′ (x).
Again, we must use the chain rule. The outside function is g(x) = log5 (x), and
1
the inside function is h(x) = x2 . By the chain rule, we get f ′ (x) =
⋅ 2x =
(ln5)x2
2x
2
=
.
2
(ln5)x
(ln5)x
We will not prove the differention formula for logarithms, but it is important to
remember it. Many of the examples involving derivatives of logarithms willl use
the chain rule or the product rule. Let’s move on to exponential formulas.
Theorem 2.2 (Derivatives for exponentials). Suppose f (x) = ax for some number
a > 1. Then f ′ (x) = (lna)ax . In particular, for a = e, if f (x) = ex , then f ′ (x) = ex .
Notice the last part of the theorem; it says that the function f (x) = ex is very
special. It is its own derivative! That means at every single point on the graph of
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ex , the tangent line at the point (a, ea ) is precisely ea . That’s pretty cool. In fact,
this is one way to define the number e. Suppose we do not know about Euler’s
constant and ask if there is a function where the slope of the graph is exactly the
y value at any given point (x, y). That is, if we have a function f (x), then for any
point (x, y) on the curve, we want f ′ (x) = f (x) = y. Let’s pick a few y values. Say
y = .5, y = 1 and y = 3. At these points, we must have slopes .5, 1 and 3 respectively.
Consider the figure below. Notice that the points must be in increasing order by
y value. (Why is that?)
y
x
Notice that the graph is beginning to look like that of an exponential graph with
positive base. That is f (x) = ax for some a > 1. That being the case, we now that
f (0) = 1 no matter what a is because a0 = 1 for any a > 1. Thus, the point (0, 1)
will be on the graph. Let’s just draw a few graphs and see what happens.
y = 1.5x
y = 2x
y = ex
y = 3x
y = 4x
y
x
We can see that the number we are looking for is e. So it happens to work out
that this special number is the key for finding a function that is its own derivative.
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Example 3. Let f (x) = e3x . Find the slope of the tangent line at x = 2.
We use the chain rule to get f ′ (x) = 3e3x . So the slope of the tangent line at
x = 2 is 3e6 .
Example 4. Let f (x) = ex ⋅ ln(x). Find f ′ (x).
Here, we have to use the product rule as we have a product of functions. Ap1
1
plying the product rule, we get f ′ (x) = ex ⋅ + ln(x) ⋅ ex = ex ( + ln(x)).
x
x