Unit2.pdf

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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
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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
h0
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
h0
f(x+h)f(x)
h= lim
h0
(x+h)2x2
h
= lim
h0
x2+ 2xh +h2x2
h= lim
h0
2xh +h2
h
= lim
h0
h(2x+h)
h= lim
h0(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.
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Unit 2 17
That is, if we recognize that a limit has the form lim
h0f(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
h0ln(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
h0f(x+h)f(x)
hevaluated at x= 2
But we know that lim
h0f(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
h0ln(x+h)ln x
h=f(x) = 1
xand so lim
h0ln(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) = exln xwe get f(x) = d
dx (ex)d
dx (ln x) = ex1
x.
The power rule:
For any constant n, the derivative of xnis nxn1. For instance, the derivative of x2is 2xand the
derivative of 1
x=x1is (1)x11=x2=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)=c0x1=c(0) = 0
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