8 Feb 2013

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Math 1225A/B

Unit 2:

Review of Introductory Calculus

Diﬀerentiation Rules and Other Calculus Techniques

(text reference: Sections 5.3 and 5.4 – mixed together

custom text pgs. 19 - 36)

c

V. Olds 2012

16 Unit 2

2 Diﬀerentiation of Exponential and Logarithmic Functions

In this section we review all the diﬀerentiation rules you learnt in your Introductory Calculus course,

mostly in the context of ﬁnding derivatives of functions involving exponentials and logarithms. We

start with the derivatives of the natural exponential and logarithmic functions. Then later we derive

formulas for the derivatives of more general exponential and logarithmic functions, i.e. with any

base b(with b > 0 and b6= 1, of course). Along the way, we will also review implicit diﬀerentiation

and logarithmic diﬀerentiation, as well as reviewing what derivatives tell us about the shape of the

graph of a function.

Derivatives of the Natural Exponential and Logarithmic Functions

Recall the Deﬁnition of Derivative.

Deﬁnition 2.1. For any function f, the derivative function f′(x) is the function given by

f′(x) = lim

h→0

f(x+h)−f(x)

h

provided this limit exists. (If the limit DNE, we say that fis not diﬀerentiable.)

Remember: This is the formula for the instantaneous rate of change in f(x) – which is the slope of

the tangent line to y=f(x). So if we evaluate f′(x) at x=a, we get the slope of the tangent line

to y=f(x) at x=a.

You probably remember doing things like the following example.

Example 2.1.Use the deﬁnition of derivative to ﬁnd f′(x) if f(x) = x2.

Solution:

f′(x) = lim

h→0

f(x+h)−f(x)

h= lim

h→0

(x+h)2−x2

h

= lim

h→0

x2+ 2xh +h2−x2

h= lim

h→0

2xh +h2

h

= lim

h→0

h(2x+h)

h= lim

h→0(2x+h) = 2x

Unfortunately, applying the deﬁnition of derivative to ﬁnd the derivative of f(x) = exor the deriva-

tive of f(x) = ln xrequires mathematics that is beyond the scope of this course, to evaluate the

limits. Never mind that. We will simply accept the following simple rules:

Rule: d

dx (ex) = exand Rule: d

dx (ln x) = 1

x

Notice: We said that ewas the base such that y=excrosses the y-axis with slope 1. For f(x) = ex,

the rule says that f′(x) = ex, so when x= 0 we get f′(0) = e0= 1. As promised, the slope (of the

tangent line) as the curve crosses the y-axis (i.e. at x= 0) is 1.

Now that we know the 2 facts mentioned above (the Rules), we can use them to recognize a

certain limit as giving the derivative of some particular function, at some value, according to the

deﬁnition of derivative. That allows us to ﬁnd the value of the speciﬁed limit without having to

evaluate it, by instead evaluating the derivative function.

Unit 2 17

That is, if we recognize that a limit has the form lim

h→0f(a+h)−f(a)

hfor some function f(x)

and some value a, then we know that this limit is giving f′(x) when evaluated at x=a, so we can

ﬁnd the value of the limit simply by evaluating f′(a). We ﬁnd the derivative of the function f(x)

which we see being used, and then evaluate it at the particular avalue we see in the limit.

Example 2.2.Evaluate lim

h→0ln(2 + h)−ln 2

h.

Solution:

Looking at this limit, we see that we can’t evaluate it in the usual way, because we don’t know how

to get the hout of the logarithm, and we can’t cancel the hin the denominator while the hin the

numerator is inside a logarithm. However, we do see a limit which has a familiar form. First of

all, we are taking the limit as hgoes to 0, and the denominator is just h. Also, in the numerator,

we see something being done ﬁrst to 2 + hand then to 2. The thing that’s being done to each of

these is “taking the natural logarithm of”. And the numerator is the diﬀerence of those two natural

logarithms.

This tells us that something’s being done with the function that takes the natural logarithm of

a number. That is, we’ve got the function f(x) = ln x. Expressed in terms of that function, the

numerator of the limit is f(2 + h)−f(2). That is, we’re evaluating f(x+h)−f(x) at the particular

value x= 2. And in fact using f(x) = ln x, we can express the whole limit as

lim

h→0f(x+h)−f(x)

hevaluated at x= 2

But we know that lim

h→0f(x+h)−f(x)

h=f′(x), and so we simply need to evaluate f′(2) to ﬁnd

the value of the given limit. That is, since we know that for f(x) = ln xthe derivative function is

f′(x) = 1

x, we have:

lim

h→0ln(x+h)−ln x

h=f′(x) = 1

xand so lim

h→0ln(2 + h)−ln 2

h=f′(2) = 1

2

Of course, we don’t need to use the deﬁnition of derivative every time we want to evaluate a

derivative. You learned a variety of diﬀerentiation rules, including: the constant multiplier rule, the

sum and diﬀerence rules, the power rule, the product rule and the quotient rule. Let’s brieﬂy review

what those rules say and how they are used.

•The constant multiplier rule:

For any diﬀerentiable function fand any constant c, the derivative of ctimes f(x) is ctimes

f′(x). For instance, for f(x) = 2exwe get f′(x) = 2 d

dx (ex)= 2ex.

•The sum and diﬀerence rules:

For any diﬀerentiable functions fand g, the derivative of h(x) = f(x)±g(x) is h′(x) =

f′(x)±g′(x). For instance, for f(x) = x+exwe get f′(x) = d

dx (x) + d

dx (ex) = 1 + ex.

Similarly, for f(x) = ex−ln xwe get f′(x) = d

dx (ex)−d

dx (ln x) = ex−1

x.

•The power rule:

For any constant n, the derivative of xnis nxn−1. For instance, the derivative of x2is 2xand the

derivative of 1

x=x−1is (−1)x−1−1=−x−2=−1

x2. Of course, it is this rule which also gives

us d

dx (x) = 1 and d

dx (c) = 0 for any constant c. We have d

dx (x) = d

dx (x1) = 1x0= 1(1) = 1.

Likewise, for any constant c,c=cx0and so (using the constant multiplier rule as well as the

power rule) we have

d

dx(c) = d

dx (cx0) = cd

dx (x0)=c0x−1=c(0) = 0