Bob Brown
CCBC Dundalk
Math 251 Calculus 1 Chapter 3, Section 2 Completed
Review of the Limit Definition of the Derivative
Write the limit definition of the derivative function:
f (x) =
Derivative of a Constant Multiple of a Function
Theorem: If f is a differentiable function and if c is a constant, then

We also use the notation c  f (x) =
d
c  f (x) =
dx
Exercise 1: Prove this Theorem using the definition of the derivative.
c  f (x)
= lim
= lim
h0
h0
=
Exercise 2: Illustrate the Theorem above with a geometric example.
y = f(x)
Estimate the derivative of f at x = -2.
Estimate the derivative of f at x = 3.
y = g(x) = 4f(x)
Estimate the derivative of g at x = -2.
Estimate the derivative of g at x = 3.
1
Bob Brown
CCBC Dundalk
Math 251 Calculus 1 Chapter 3, Section 2 Completed
Derivative of the Sum or Difference of Functions
Theorem: If f and g are differentiable functions, then
 f ( x)  g ( x) =
d
 f ( x)  g ( x) =
dx
Exercise 3: Prove the Difference Rule using the definition of the derivative. (The proof
for the Sum Rule is found on page 130 in the text.)
 f ( x)  g ( x) =
lim
= lim
h0
h0
Derivative of a Power Function (The Power Rule)
Def: A power function is a function of the form
, where
Theorem: The derivative function of f ( x)  x n is
Note: The proof and/or hints for a proof of the Power Rule for various conditions on n
can be found on pages 127, 143, and 173 in the text.
Derivative of y = c and of y = x
The derivative of the constant function f(x) = c is
The derivative of the function g(x) = x is
f (x) =
That is,
d
(c ) =
dx
Use the Power Rule to show this fact.
g (x ) =
That is,
d
(x ) =
dx
Use the Power Rule to show this fact.
f(x) = c =
g(x) = x =
f (x) =
g (x ) =
Use the limit definition to prove this fact.
Use the limit definition to prove this fact.
2
Bob Brown
CCBC Dundalk
Math 251 Calculus 1 Chapter 3, Section 2 Completed
3
Exercise 4: Use the Power Rule to determine each derivative function.
(i) y = x123
(iii) y 
1
x4
(ii) y = x12.3
(iv) y 
1
4
x
(v) y = x3
(vi) y = 3x
(vii) y = xπ
(viii) y = xln(2)
Derivative Functions of Polynomials and of Some Other Functions
We can now determine the derivative function of a function expressed (or expressable) as
the sum or difference of power functions, including polynomials. (Note that only the
function in (i) below is a polynomial.)
Exercise 5: Determine the derivative function.
(i) y  7 x 3  5x 2  4 x
(iii) y 
x9  6x5  x 2
x4
(ii) y  x 2 
(iv) y 
1
2x
x2 1
3
x
Bob Brown
CCBC Dundalk
Math 251 Calculus 1 Chapter 3, Section 2 Completed
4
Exercise 6: What is derivative function of f(x) = sin(x)? To answer this question, consider the
graph of y = sin(x) on the interval 0  x  2 drawn below. As we did in Handout 3.1, draw the
tangent line at various points, estimate the slope, and sketch f (x ) .
x
0

2

3
2
2
f (x)
Draw f (x) below.
Conclusion: If f(x) = sin(x), then f (x) =
Exercise 7: What is derivative function of g(x) = cos(x)? To answer this question, consider the
graph of y = cos(x) on the interval 0  x  2 drawn below. As we did in Handout 3.1, draw
the tangent line at various points, estimate the slope, and sketch g (x ) .
x
0

2

g (x )
Draw g (x ) below.
Conclusion: If g(x) = cos(x), then g (x ) =
3
2
2
Bob Brown
CCBC Dundalk
Math 251 Calculus 1 Chapter 3, Section 2 Completed
The Derivative of an Exponential Function
Exercise 8: Consider f(x) = 2x. Write the limit definition of each of the following, and
use the table feature of the TI to estimate the value of each limit to four decimal places.
f (0) = lim
h 0
f (0  h)  f (0)
= lim
h0
h
f (1) = lim
h0
f ( 2) =
So,
f (0) 
=
 20 =
 f (0)
f (1) 
=
 21 =
 f (1)
f ( 2) 
=
 22 =
 f ( 2)
 2x =
 f (x )
f (x) 
We recognize the decimal 0.6931 as an approximation of
We conclude that if f(x) = 2x, then f (x) =
Note that the derivative function of f(x) = 2x is merely a constant multiple of f(x) = 2x.
The constant multiple, in this case, is
If f(x) = 3x, then f (x) =
. Note that the derivative function of f(x) = 3x is
merely a constant multiple of f(x) = 3x. The constant multiple, in this case, is
Recall: If one function is a constant multiple times another function, then its graph is a
vertical stretch or shrink (depending on whether the constant multiple is greater than 1 or
less than 1) of the graph of the other function.
5
Bob Brown
CCBC Dundalk
Math 251 Calculus 1 Chapter 3, Section 2 Completed
6
Question: Is there an exponential function such that the constant multiple in the formula
for the derivative function is 1? That is to say, is there an exponential function that is
equal to its derivative function?
Exercise 9: Let f(x) = ax. Write the limit definition of the derivative function, and
compute the value of a for which f(x) = f (x) .
f (x) = lim
h 0
f ( x  h)  f ( x )
= lim
h0
h
Derivative Function of an Exponential Function y = ax (where a > 0)
 
d x
a =
dx
(see p.158 in the text)
“Special”case worth nothing:
 
d x
e =
dx
Bob Brown
CCBC Dundalk
Math 251 Calculus 1 Chapter 3, Section 2 Completed
7
Exercise 10: Determine the derivative function of each of the following functions.
(ii) y  2  5 x
(i) y = 4x
x
(iii) y = -6π + 3e
2
(iv) y   
3
(v) y = (ln(2))x
(vi) y = 3x + x3
(vii) y = (2e)x
(viii) y = e3x
x
x
Bob Brown
CCBC Dundalk
Math 251 Calculus 1 Chapter 3, Section 2 Completed
Average Velocity and Instantaneous Velocity of a Rocket
Exercise 11: A rocket is launched vertically from a submarine 200 feet below the surface of
the ocean with an initial velocity of 1500 feet per second. (For simplicity, we will ignore water
friction and air friction. In reality, neither can be ignored.) The following table and graph
represent y = the height in feet of the rocket above the level of the water as a function of
x = time in seconds since the beginning of the launch. (The graph does not indicate the path of
the rocket.) The equation which relates y and x is y = -16x2 + 1500 x – 200. Note that the
x-axis is scaled in increments of 10, and the y-axis is scaled in increments of 5000.
(i) Using the table, compute the average velocity of
the rocket between x = 20 sec. and x = 50 sec.
Draw the corresponding secant line.
(ii) Compute the average velocity of the rocket
between x = 20 sec. and x = 25 sec. Draw the
corresponding secant line.
(iii) The instantaneous velocity of the rocket at x = 20 seconds is the slope of the tangent line to the graph
at x = 20. Compute this value exactly (using derivative rules, not by estimating a slope geometrically.)
(iv) What is the velocity (“velocity” means “instantaneous velocity” unless otherwise qualified) of the
rocket at the moment in its descent that it is 10,000 feet above the surface of the water?
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