Section 3–2 Examples
Basic Derivative Rules
The Derivative of a Constant Rule
Theorem: The derivative of a constant is zero
(c will stand for a constant in this course)
If y = c then y′ = 0
(
Note: There are many expressions that are constants. 3π , e, lne, ln2, e 2, sin(2π ), sin2 θ + cos2 θ
)
Example 1
Find y ′ if y = −4
Example 2
Find f ′ if f (x) = 3e
Example 3
Find f ′(x) if f (x) = 2π
y′ = 0
f ′(x) = 0
f ′(x) = 0
The Power Rule
Theorem:
If a,n ∈ ℜ and
f(x) = ax n
then f ′(x) = a• n• x n−1
Note: f(x) must be EXACTLY a • xn
Example 1
Example 2
Example 3
Find y ′ if y = 9x
Find y ′ if y = 6x 3
Find y ′ if y = 4x −2
y ′ = 6• 3• x 3− 1
y ′ = 4 • −2 • x −2−1
y ′ = 9• 1• x1−1
y ′ = 18• x 2
y ′ = −8 • x −3
−8
y′ = 3
x
y ′ = 9• x 0
y′ = 9
Note: 9x = 9x1
Example 4
Find y ′ if y = 15 3 x 5 = 15x 5 3
Example 5
3
Find y ′ if y = −9 x 2 = −9 x 2 3
5
−1
5
y ′ = 15• • x 3
3
2
−1
2
y ′ = −9 • • x 3
3
y ′ = 25• x 2 3
y ′ = −6 • x −1 3
−6
y′ = 1 3
x
4
y′ = 3
x
3
y ′ = 25 x 2
Math 400 3–2 Basic Derivative Rules
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If y =
c
then y = cx − n
xn
A fraction with a monomial term n the denominator may be changed into the power rule.
Example 1
Example 2
Find y ′ if y = 3
x
Find y ′ if y =
y = 6x −3
−1
y ′ = 10• • x
2
y ′ = −5 • x −3 2
−5
y′ = 3 2
x
−5
y′ =
x x
6
y ′ = 6• (−3) • x −3−1
y ′ = −18 • x −4
−18
y′ = 4
x
10
= 10x −1/2
x
−1
−1
2
The Constant Multiple Rule
The derivative of a constant times a function is the constant times the derivative of that function
If c is a real number then
d
[c • f (x) ]
dx
d
= c•
f (x)
dx
= c • f ′(x)
Example 1
Example 2
Find y ′ if y = 5x −3
y ′ = 5•
Find y ′ if y =
[ ]
d −3
x
dx
(
y ′ = 5 −3 • x −3−1
)
y ′ = −5 • x −3 2
−5
y′ = 3 2
x
−5
y′ =
x x
−15
x4
Math 400 3–2 Basic Derivative Rules
)
d −1/2
x
dx
−1 ⎞
⎛
−1
−1
y ′ = 10• ⎜ • x 2 ⎟
⎜2
⎟
⎝
⎠
y ′ = 10•
y ′ = −15 • x −4
y′ =
(
10
= 10x −1/2
x
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The Sum and Difference Rule
The derivative of the sum or difference of two (or more) differentiable functions
is the derivative of each of the seperate functions:
d
[ f (x) ± g(x) ± h(x) ± ......]
dx
= f ′ (x) ± g′(x) ± h′ (x) ± ...
NOTE: This rule cannot be used for the PRODUCT of 2 functions. It can only be used with
the sum of functions
Example 1
Example 2
Find y ′ if y = 6x 3 + 6x 2 − 5x
Find y ′ if y =
y ′ = 18x 2 + 12x − 5
1 6
x − x 3 − 3x + 8
3
y ′ = 2x 5 − 3x 2 − 3
A fraction with a polynomial in the numerator and a
MONOMIAl term n the denominator may be changed into the power rule.
Example 1
Find y ′ if y =
y=
15x 5 − 12x 4 + 9x 3
3x 3
15x 5
12x 4
9x 3
−
+
3x 3
3x 3
3x 3
y = 5x 2 − 4x + 3
y ′ = 10x − 4
Example 2
Find y ′ if y =
y=
4x 2
1/2
2x
−
4x 2 − 16x
2 x
16x
2x 1/2
y = 2x 3/2 − 8x 1/2
y ′ = 3x 1/2 − 4 x −1/2
4
y′ = 3 x −
x
Math 400 3–2 Basic Derivative Rules
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A binomial raised to a power
Find f ′ if f (x) = (3x + 5) 2
The power rule at this point in the chapter states the derivative of a• x n
This rule would fit the expression x 2 but does not fit the expression (3x + 5) 2
If you FOIL out (3x + 5) 2 then you get 3 expressions that do fit the rules of this section
Find f ′ if f (x) = (3x + 5) 2
= (3x + 5) • ( 3x + 5)
f (x) = 9x 2 + 30x + 25
f ′(x) = 18x + 30
A binomial raised to a power can be changed into the power rule by FOILing
Example
Find y ′ if y = (3x 2 − 2x) 2
FOIL the expression
= 9x 4 − 12x 3 + 4 x 2
y ′ = 36x 3 − 36x 2 + 8x
Math 400 3–2 Basic Derivative Rules
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The Derivative of the Exponential Function f (x) = ae x
If f(x) = ae x
then f ′(x) = ae x
Example 1
Note: f(x) must be EXACTLY f(x) = a e x
Example 2
Find y ′ if y = 6e x
Find y ′ if y = 4e x − 4x 2 + 3
y ′ = 6e x
y ′ = 4e x − 8x
The Derivative of f(x) = aln(x)
If f(x) = aln(x)
a
then f ′(x) =
x
Note: f(x) must be EXACTLY f(x) = aln(x)
Example 1
Example 2
Find y ′ if y = ln x 3
y = 3ln x
3
y′ =
x
Find y ′ if y = 6ln(x)
y′ =
6
x
The Derivative of f(x) = a x
The Derivative of f(x) = a log b (x)
If f(x) = a logb (x)
a
then f ′(x) =
ln(b) x
If f(x) = a x
then f ′(x) = ln(a)• a x
Note: f(x) must be EXACTLY
f(x) = alog b (x)
Note: f(x) must be EXACTLY
f(x) = a x
Example 1
Example 1
Find y ′ if y = 3x
Find y ′ if y = 6log(x)
y′ =
y ′ = ln(3)• 3 x
6
ln(10)x
Math 400 3–2 Basic Derivative Rules
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The Derivative of the trigonometric functions
d
[sin x ] = cos x
dx
d
[cos x ] = − sin x
dx
d
[tan x ] = sec 2 x
dx
d
[csc x ] = − csc x • cot x
dx
d
[sec x ] = sec x • tan x
dx
d
[cot x ] = − csc 2 x
dx
Note: Each function must be exactly
a• sinx or a • cos x or a • tan x or a • csc x or a• sec x or a• cot x
Example 1
Example 2
Example 3
Find y ′ if y = 6sin x
Find f ′(x) if f (x) = −3• cosx
Find y ′ if y = 4tan x
y ′ = 6cosx
f ′(x) = −3(− sin x )
f ′(x) = 3sin x
y ′ = 4 sec2 x
Example 4
Example 5
Example 6
Find y ′ if y = 2csc x
Find f ′(x) if f (x) = −3• sec x
Find y ′ if y = −5 cot x
y ′ = −2csc x • cot x
f ′(x) = −3 • sec x • tan x
y ′ = 5csc2 x
Example 7
Find y ′ if y =
−2 6
x + 5e x − 3sin x − 5cos x + 4tan x + sec x
3
y ′ = −4 x 5 + 5e x − 3cos x + 5sin x + 4sec 2 x + sec x • tan x
Math 400 3–2 Basic Derivative Rules
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The Second Derivative
The notation for the first derivative is shown below. The first 3 notations are the most commonly used.
dy
d
[ f (x)] First Derivative
y′
f ′(x)
dx
dx
The notation for the second derivative is shown below. The first 3 notations are the most commonly used.
Second Derivative
y ′′
f ′′(x)
d2y
dx 2
d2
dx 2
[ f (x)]
The Second Derivative is the derivative of the first derivative. To find the second derivative of
f (x) find the first derivative f ′(x) and then take the derivative of the first derivative,
Example 1
Example 2
Find f ′′ if f (x) = −3x −2 + e x
Find f ′′ if f (x) = 3x 4 + 5x 3
The first derivative of f (x) is
The first derivative of f (x) is
f ′ (x) = 6x −3 + e x
f ′ (x) = 12x 3 + 15x 2
the derivative of the first derivative f ′ ( x)
is the second derivative f ′′ (x)
the derivative of the first derivative f ′ ( x)
is the second derivative f ′′ (x)
f ′′ (x) = 18x −4 + e x
f ′′ (x) = 36x 2 + 30x
18
f ′′ (x) = 4 + e x
x
Example 3
Example 4
Find f ′′ if f (x) = 3sin(x)
Find f ′′ if f (x) = 12 x + 8x 3/2
f (x) = 12x 1/2 + 4x 5/2
The first derivative of f (x) is
f ′ (x) = 3cos(x)
The first derivative of f (x) is
f ′ (x) = 6x −1/2 + 12x 3/2
the derivative of the first derivative f ′ ( x)
is the second derivative f ′′ (x)
f ′′ (x) = −3sin( x)
the derivative of the first derivative f ′ ( x)
is the second derivative f ′′ (x)
f ′′ (x) = −3x −3/2 + 18x1/2
f ′′ (x) =
−3
+ 18 x
x3
Math 400 3–2 Basic Derivative Rules
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Finding the slope of the line tangent to the graph of f (x) at x = c where c ∈ Reals
The slope of the line tangent to f(x) at x = c is f ′(c)
Example 1
Example 2
Find the slope of the tangent line
3
at x = 4 if f (x) = x 2
2
y ′ = 3x
y ′(4) = 12
Find the sloipe of the tangent line
π
at x =
if f (x) = sin x
6
f ′(x) = cos x
f′
slope of the tangent line
at x = 4 is 12
3
⎛π⎞
=
⎝ 6⎠
2
slope of the tangent line
at x =
π
3
is
6
2
Finding the equation of the line tangent to the graph of f(x) at x = c where c ∈ Reals
The point on the graph of f(x) line is (c, f (c) )
The slope of the line tangent to f(x) at x = c is f ′(c)
Using the point slope form of the equation of a line y − y1 = m( x − x 1 )
the equation of the line tangent to the graph of f(x) at x = c is
Example 1
Find the equation of the tangent line
3
at x = 4 if f (x) = x 2
2
the line is tanget to the function
at the point (4,24)
y ′ = 3x
y ′(4) = 12
Example 2
Find the equation of the tangent line
π
at x =
if f (x) = sin x
6
the line is tanget to the function
Math 400 3–2 Basic Derivative Rules
⎛ π 1⎞
,
⎝ 6 2⎠
at the point
f ′(x) = cos x
f′
y − f (c) = f ′(c)( x − c)
y − 24 = 12(x − 4)
or
y = 12x − 24
y − f (c) = f ′(c)( x − c)
3
⎛π⎞
=
⎝ 6⎠
2
y − f (c) = f ′(c)( x − c)
y−
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1
3⎛
π
=
x− ⎞
⎝
2
2
6⎠
©2013 Eitel
Finding where the function has a tangent line with a given slope
A function has a slope of m where f ′(c) = m
Example 1
Example 2
Find all x values where f (x) = 4 x 2 − 2x
has a slope of 4
f ′(x) = 8x − 2
Find all x values where f (x) =
1 3
x − 2x
3
has a slope of 7
f ′(x) = x 2 − 2
set f ′ (x) = 4 and solve for x
8x − 2 = 4
8x = 6
set f ′ (x) = 7 and solve for x
x = 3/4
x2 − 9= 0
(x − 3)(x + 3) = 0
x2 − 2= 7
x = 3 or − 3
Finding where the function has a positive or negative slope
A function has a positive slope where f ′(x) > 0
A function has a positive slope where f ′(x) < 0
Example 1
Example 2
Find all x values where f (x) = 6x 2 − 24x
has a positive slope
f ′(x) = 12x − 24
Find all x values where f (x) = −3x 2 − 18x
has a negitive slope
f ′(x) = −6x − 18
set f ′ (x) > and solve for x
12x − 24 > 0
12x > 24
x >2
set f ′ (x) < 0 and solve for x
−6 x − 18 < 0
−6x > 18
x < −3
ℜ|x> 2
or in interval notation (2,∞)
ℜ | x < −3
or in interval notation(− ∞, −3)
Math 400 3–2 Basic Derivative Rules
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Finding where the function has horizontal tangent lines.
A function has horizontal tangent lines at all values of x where f ′(x) = 0
Example 1
Example 2
Find all x values where f (x) = x 3 − 27x
has a horizontail tangent line.
Find all x values where f (x) = sin x
has a horizontail tangent line.
f ′(x) = cos x
f ′(x) = 3x 2 − 27
set f ′ (x) = 0 and solve for x
set f ′ (x) = 0 and solve for x
cos x = 0
3x 2 − 27 = 0
3(x + 3)(x − 3) = 0
π
+π n
2
where n ∈ integers
ℜ|x=
x = 3 or − 3
Math 400 3–2 Basic Derivative Rules
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The Position Function s(t) of a falling body
The position of a free falling body, without regard to wind resistance, is represented by the equation
1
s(t) = g• t 2 + v 0 t + s0
2
where s0 is the initial height of the object. g is the acceleration due to gravity. On earth thiis acceleration is
–32 feet per second squared or –9.8 meters per second squared. v 0 is the initial velocity. If the object is
dropped then the initial v 0 is 0. If the object is shot upwards then v 0 is positive and if the object is
thrown downwards then is v 0 is negative.
Example 1
A ball is thrown downwards at a velocity of 4 feet per second from a height of 120 feet above the
ground.
s(t) = −16 t 2 − 4t + 120
When t = 2 the ball is s(2) = −16(2) 2 − 4(2) + 120 = 48 48 feet above the ground
Example 2
A ball is thrown upwards at a velocity of 10 feet per second from a height of 40 feet above the ground.
s(t) = −16 t 2 + 10t + 40
When t = 1 the ball is s(1) = −16(1) 2 + 10(1) + 40 = 34 34 feet above the ground.
Example 3
A ball is thrown upwards at a velocity of 16 feet per second from a height of 32 feet above the ground.
Find when the ball hits the ground.
s(t) = −16 t 2 + 16t + 32
If s(t) = −16 t 2 + 16t + 32
then the ball hits the ground when s(t) = 0
−16 t 2 + 16t + 32 = 0
−16(t 2 − t − 2) = 0
−16(t + 1)(t − 2)
t = −1 sec or 2 sec
time must be > 0 so
the ball hits the ground when
t = 2 sec
Math 400 3–2 Basic Derivative Rules
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The Velocity Function v(t)
The Velocity Function v(t) is the derivative of the position function.
1
g • t 2 + v 0t + s0
2
ds
then v(t) =
dt
v(t) = g• t + v 0
If s(t) =
Example 1
A ball is thrown upwards at a velocity of 10 feet per second from a height of 40 feet above the ground.
What is the velocity when t = 3 sec.
s(t) = −16 t 2 + 10t + 40
If s(t) = −16 t 2 + 10t + 40
than v(t) = −32t 2 + 10
If t = 3 sec then
v(3) = −32(3) 2 + 10 = −278 ft sec
Example 2
A ball is thrown downwards at a velocity of 4 feet per second from a height of 120 feet above the
ground. Part A: Find the Velocity Function v(t) . Part B: Find when the velocity will be –128 feet per
second.
If s(t) = −16 t 2 − 4t + 120
than v(t) = −32t 2 − 4
If v(t) = −32t 2 − 4
than v(t) = −132 f s2 when
−32t 2 − 4 = −132
−32t = −128
t = 2 sec
The Acceleration Function a(t)
The Acceleration Function a(t) is the derivative of the Velocity Function v(t) .
Math 400 3–2 Basic Derivative Rules
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