3.3 differentiation rules -...
TRANSCRIPT
§ 3.3 Differentiation Rules
Constant Functions
What does a constant function look like?
x
y
y = 2
What is the slope?
Constant Functions
What does a constant function look like?
x
y
y = 2
What is the slope?
Constant Functions
What does a constant function look like?
x
y
y = 2
What is the slope?
Constant Functions
What does a constant function look like?
x
y
y = 2
y = 0
Constant Function Ruleddx c = 0 for all c ∈ R.
Constant Function Rule
Why?
f ′(x) = limh→0
f (x + h)− f (x)
h
= limh→0
c− ch= 0
Constant Function Rule
Why?
f ′(x) = limh→0
f (x + h)− f (x)
h
= limh→0
c− ch= 0
Constant Function Rule
Why?
f ′(x) = limh→0
f (x + h)− f (x)
h
= limh→0
c− ch
= 0
Constant Function Rule
Why?
f ′(x) = limh→0
f (x + h)− f (x)
h
= limh→0
c− ch= 0
Linear Functions
What is the slope of a linear function?
x
y
y = 12 x− 1
What is the slope?
Linear Functions
What is the slope of a linear function?
x
y
y = 12 x− 1
What is the slope?
Linear Functions
What is the slope of a linear function?
x
y
y = 12 x− 1
What is the slope?
Linear Functions
What is the slope of a linear function?
x
y
y = 12 x− 1
y = 12
Quadratic Functions
What does a quadratic function look like?
x
y
Quadratic Functions
What does a quadratic function look like?
x
y
Quadratic Functions
What is the slope at a point?
x
y
y = x2 − 1
y = 2x
Quadratic Functions
What is the slope at a point?
x
y
y = x2 − 1
y = 2x
Quadratic Functions
What is the slope at a point?
x
y
y = x2 − 1
y = 2x
Cubic Functions?
x
y
Cubic Functions?
x
y
Power Functions
ddx
xn = limh→0
(x + h)n − xn
h
= limh→0
xn + nxn−1h + n(n−1)2 xn−2h2 + . . . + hn − xn
h
= limh→0
nxn−1h + n(n−1)2 xn−2h2 + . . . + hn
h
= limh→0
h(nxn−1 + n(n−1)2 xn−2h + . . . + hn−1)
h
= limh→0
nxn−1 +n(n− 1)
2xn−2h + . . . + hn−1
= nxn−1
Power Functions
ddx
xn = limh→0
(x + h)n − xn
h
= limh→0
xn + nxn−1h + n(n−1)2 xn−2h2 + . . . + hn − xn
h
= limh→0
nxn−1h + n(n−1)2 xn−2h2 + . . . + hn
h
= limh→0
h(nxn−1 + n(n−1)2 xn−2h + . . . + hn−1)
h
= limh→0
nxn−1 +n(n− 1)
2xn−2h + . . . + hn−1
= nxn−1
Power Functions
ddx
xn = limh→0
(x + h)n − xn
h
= limh→0
xn + nxn−1h + n(n−1)2 xn−2h2 + . . . + hn − xn
h
= limh→0
nxn−1h + n(n−1)2 xn−2h2 + . . . + hn
h
= limh→0
h(nxn−1 + n(n−1)2 xn−2h + . . . + hn−1)
h
= limh→0
nxn−1 +n(n− 1)
2xn−2h + . . . + hn−1
= nxn−1
Power Functions
ddx
xn = limh→0
(x + h)n − xn
h
= limh→0
xn + nxn−1h + n(n−1)2 xn−2h2 + . . . + hn − xn
h
= limh→0
nxn−1h + n(n−1)2 xn−2h2 + . . . + hn
h
= limh→0
h(nxn−1 + n(n−1)2 xn−2h + . . . + hn−1)
h
= limh→0
nxn−1 +n(n− 1)
2xn−2h + . . . + hn−1
= nxn−1
Power Functions
ddx
xn = limh→0
(x + h)n − xn
h
= limh→0
xn + nxn−1h + n(n−1)2 xn−2h2 + . . . + hn − xn
h
= limh→0
nxn−1h + n(n−1)2 xn−2h2 + . . . + hn
h
= limh→0
h(nxn−1 + n(n−1)2 xn−2h + . . . + hn−1)
h
= limh→0
nxn−1 +n(n− 1)
2xn−2h + . . . + hn−1
= nxn−1
Power Functions
ddx
xn = limh→0
(x + h)n − xn
h
= limh→0
xn + nxn−1h + n(n−1)2 xn−2h2 + . . . + hn − xn
h
= limh→0
nxn−1h + n(n−1)2 xn−2h2 + . . . + hn
h
= limh→0
h(nxn−1 + n(n−1)2 xn−2h + . . . + hn−1)
h
= limh→0
nxn−1 +n(n− 1)
2xn−2h + . . . + hn−1
= nxn−1
Power Functions
Derivative of Power Functionsddx
xn = nxn−1
for all n ∈ R.
How does this work with negative numbers? Fractions?
Power Functions
Derivative of Power Functionsddx
xn = nxn−1
for all n ∈ R.
How does this work with negative numbers? Fractions?
Examples
Example
Find ddx
1x .
ddx
1x
=ddx
x−1
= −x−2
=−1x2
Examples
Example
Find ddx
1x .
ddx
1x
=
ddx
x−1
= −x−2
=−1x2
Examples
Example
Find ddx
1x .
ddx
1x
=ddx
x−1
= −x−2
=−1x2
Examples
Example
Find ddx
1x .
ddx
1x
=ddx
x−1
= −x−2
=−1x2
Examples
Example
Find ddx
1x .
ddx
1x
=ddx
x−1
= −x−2
=−1x2
Examples
Example
Find ddx√
x.
ddx√
x =ddx
x12
=12
x( 12−1)
=1
2x12
=1
2√
x
Examples
Example
Find ddx√
x.
ddx√
x =
ddx
x12
=12
x( 12−1)
=1
2x12
=1
2√
x
Examples
Example
Find ddx√
x.
ddx√
x =ddx
x12
=12
x( 12−1)
=1
2x12
=1
2√
x
Examples
Example
Find ddx√
x.
ddx√
x =ddx
x12
=12
x( 12−1)
=1
2x12
=1
2√
x
Examples
Example
Find ddx√
x.
ddx√
x =ddx
x12
=12
x( 12−1)
=1
2x12
=1
2√
x
Examples
Example
Find ddx√
x.
ddx√
x =ddx
x12
=12
x( 12−1)
=1
2x12
=1
2√
x
Constant Multiple Rule
Example
Find ddx 3x2.
ddx
3x2 = limh→0
3(x + h)2 − 3x2
h
= limh→0
3((x + h)2 − x2)
h
= limh→0
3(x2 + 2xh + h2 − x2)
h
= limh→0
3h(2xh + h2)
h= lim
h→03(2x + h)
= 3(2x) = 6
Constant Multiple Rule
Example
Find ddx 3x2.
ddx
3x2 = limh→0
3(x + h)2 − 3x2
h
= limh→0
3((x + h)2 − x2)
h
= limh→0
3(x2 + 2xh + h2 − x2)
h
= limh→0
3h(2xh + h2)
h= lim
h→03(2x + h)
= 3(2x) = 6
Constant Multiple Rule
Example
Find ddx 3x2.
ddx
3x2 = limh→0
3(x + h)2 − 3x2
h
= limh→0
3((x + h)2 − x2)
h
= limh→0
3(x2 + 2xh + h2 − x2)
h
= limh→0
3h(2xh + h2)
h= lim
h→03(2x + h)
= 3(2x) = 6
Constant Multiple Rule
Example
Find ddx 3x2.
ddx
3x2 = limh→0
3(x + h)2 − 3x2
h
= limh→0
3((x + h)2 − x2)
h
= limh→0
3(x2 + 2xh + h2 − x2)
h
= limh→0
3h(2xh + h2)
h= lim
h→03(2x + h)
= 3(2x) = 6
Constant Multiple Rule
Example
Find ddx 3x2.
ddx
3x2 = limh→0
3(x + h)2 − 3x2
h
= limh→0
3((x + h)2 − x2)
h
= limh→0
3(x2 + 2xh + h2 − x2)
h
= limh→0
3h(2xh + h2)
h
= limh→0
3(2x + h)
= 3(2x) = 6
Constant Multiple Rule
Example
Find ddx 3x2.
ddx
3x2 = limh→0
3(x + h)2 − 3x2
h
= limh→0
3((x + h)2 − x2)
h
= limh→0
3(x2 + 2xh + h2 − x2)
h
= limh→0
3h(2xh + h2)
h= lim
h→03(2x + h)
= 3(2x) = 6
Constant Multiple Rule
Example
Find ddx 3x2.
ddx
3x2 = limh→0
3(x + h)2 − 3x2
h
= limh→0
3((x + h)2 − x2)
h
= limh→0
3(x2 + 2xh + h2 − x2)
h
= limh→0
3h(2xh + h2)
h= lim
h→03(2x + h)
= 3(2x) = 6
Constant Multiples
Constant Multiple Rule
For any differentiable function y = f (x),
ddx
cf (x) = cddx
f (x) = cf ′(x)
Sums and Differences
Sum and Difference Ruleddx
[f (x)± g(x)] = f ′(x)± g′(x)
limh→0
f (x + h)± g(x + h)− (f (x)± g(x))
h
= limh→0
f (x + h)− f (x)
h± g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
h± lim
h→0
g(x + h)− g(x)
h= f ′(x)± g′(x)
Sums and Differences
Sum and Difference Ruleddx
[f (x)± g(x)] = f ′(x)± g′(x)
limh→0
f (x + h)± g(x + h)− (f (x)± g(x))
h
= limh→0
f (x + h)− f (x)
h± g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
h± lim
h→0
g(x + h)− g(x)
h= f ′(x)± g′(x)
Sums and Differences
Sum and Difference Ruleddx
[f (x)± g(x)] = f ′(x)± g′(x)
limh→0
f (x + h)± g(x + h)− (f (x)± g(x))
h
= limh→0
f (x + h)− f (x)
h± g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
h± lim
h→0
g(x + h)− g(x)
h= f ′(x)± g′(x)
Sums and Differences
Sum and Difference Ruleddx
[f (x)± g(x)] = f ′(x)± g′(x)
limh→0
f (x + h)± g(x + h)− (f (x)± g(x))
h
= limh→0
f (x + h)− f (x)
h± g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
h± lim
h→0
g(x + h)− g(x)
h
= f ′(x)± g′(x)
Sums and Differences
Sum and Difference Ruleddx
[f (x)± g(x)] = f ′(x)± g′(x)
limh→0
f (x + h)± g(x + h)− (f (x)± g(x))
h
= limh→0
f (x + h)− f (x)
h± g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
h± lim
h→0
g(x + h)− g(x)
h= f ′(x)± g′(x)
An Example
Example
Find f ′(x) for f (x) = −2x2 + 4x3 − 3x .
f ′(x) = − 4x + 12x2 +3x2
An Example
Example
Find f ′(x) for f (x) = −2x2 + 4x3 − 3x .
f ′(x) =
− 4x + 12x2 +3x2
An Example
Example
Find f ′(x) for f (x) = −2x2 + 4x3 − 3x .
f ′(x) = − 4x
+ 12x2 +3x2
An Example
Example
Find f ′(x) for f (x) = −2x2 + 4x3 − 3x .
f ′(x) = − 4x + 12x2
+3x2
An Example
Example
Find f ′(x) for f (x) = −2x2 + 4x3 − 3x .
f ′(x) = − 4x + 12x2 +3x2
Product Rule
Let ∆f = f (x + h)− f (x) and ∆g = g(x + h)− g(x).
f (x)
g(x)
∆f
∆g
Product Rule
Let ∆f = f (x + h)− f (x) and ∆g = g(x + h)− g(x).
f (x)
g(x)
∆f
∆g
Product Rule
Let ∆f = f (x + h)− f (x) and ∆g = g(x + h)− g(x).
f (x)
g(x)
∆f
∆g
Product Rule
Let ∆f = f (x + h)− f (x) and ∆g = g(x + h)− g(x).
f (x)
g(x)
∆f
∆g
Product Rule
Let ∆f = f (x + h)− f (x) and ∆g = g(x + h)− g(x).
f (x)
g(x)
∆f
∆g
Product Rule
Consider f (x + h)g(x + h).
f (x + h)f (x)
g(x + h)
g(x)
f (x)g(x) g(x)∆f
f (x)∆g ∆f ∆g
Product Rule
Consider f (x + h)g(x + h).
f (x + h)f (x)
g(x + h)
g(x)
f (x)g(x) g(x)∆f
f (x)∆g ∆f ∆g
Product Rule
Consider f (x + h)g(x + h).
f (x + h)f (x)
g(x + h)
g(x)
f (x)g(x)
g(x)∆f
f (x)∆g ∆f ∆g
Product Rule
Consider f (x + h)g(x + h).
f (x + h)f (x)
g(x + h)
g(x)
f (x)g(x) g(x)∆f
f (x)∆g ∆f ∆g
Product Rule
Consider f (x + h)g(x + h).
f (x + h)f (x)
g(x + h)
g(x)
f (x)g(x) g(x)∆f
f (x)∆g
∆f ∆g
Product Rule
Consider f (x + h)g(x + h).
f (x + h)f (x)
g(x + h)
g(x)
f (x)g(x) g(x)∆f
f (x)∆g ∆f ∆g
As h→ 0
As h→ 0
As h→ 0
The Product Rule
ddx
[f (x)g(x)] = limh→0
f (x + h)g(x + h)− f (x)g(x)
h
= limh→0
(∆fh
g(x) + f (x)∆gh
+∆f ∆g
h
)= f ′(x)g(x) + g′(x)f (x) + lim
h→0
(∆fh
∆gh
h)
= f ′(x)g(x) + g′(x)f (x)
The Product Rule
ddx
[f (x)g(x)] = limh→0
f (x + h)g(x + h)− f (x)g(x)
h
= limh→0
(∆fh
g(x) + f (x)∆gh
+∆f ∆g
h
)
= f ′(x)g(x) + g′(x)f (x) + limh→0
(∆fh
∆gh
h)
= f ′(x)g(x) + g′(x)f (x)
The Product Rule
ddx
[f (x)g(x)] = limh→0
f (x + h)g(x + h)− f (x)g(x)
h
= limh→0
(∆fh
g(x) + f (x)∆gh
+∆f ∆g
h
)= f ′(x)g(x) + g′(x)f (x) + lim
h→0
(∆fh
∆gh
h)
= f ′(x)g(x) + g′(x)f (x)
The Product Rule
ddx
[f (x)g(x)] = limh→0
f (x + h)g(x + h)− f (x)g(x)
h
= limh→0
(∆fh
g(x) + f (x)∆gh
+∆f ∆g
h
)= f ′(x)g(x) + g′(x)f (x) + lim
h→0
(∆fh
∆gh
h)
= f ′(x)g(x) + g′(x)f (x)
The Product Rule
The Product Ruleddx f (x)g(x) = f (x)g′(x) + g(x)f ′(x)
Alternate StatementIf u = f (x) and v = g(x), then
ddx
uv = uv′ + vu′
The Product Rule
The Product Ruleddx f (x)g(x) = f (x)g′(x) + g(x)f ′(x)
Alternate StatementIf u = f (x) and v = g(x), then
ddx
uv = uv′ + vu′
An Example
Example
Find f ′(x) for f (x) = x2(x− 4)5.
f ′(x) = x2 ddx
(x− 4)5 + (x− 4)5 ddx
x2
= x2(5(x− 4)4) + (x− 4)5(2x)
= x(x− 4)4(5x + (x− 4)(2))
= x(x− 4)4(5x + 2x− 8)
= x(x− 4)4(7x− 8)
An Example
Example
Find f ′(x) for f (x) = x2(x− 4)5.
f ′(x) = x2 ddx
(x− 4)5 + (x− 4)5 ddx
x2
= x2(5(x− 4)4) + (x− 4)5(2x)
= x(x− 4)4(5x + (x− 4)(2))
= x(x− 4)4(5x + 2x− 8)
= x(x− 4)4(7x− 8)
An Example
Example
Find f ′(x) for f (x) = x2(x− 4)5.
f ′(x) = x2 ddx
(x− 4)5 + (x− 4)5 ddx
x2
= x2(5(x− 4)4) + (x− 4)5(2x)
= x(x− 4)4(5x + (x− 4)(2))
= x(x− 4)4(5x + 2x− 8)
= x(x− 4)4(7x− 8)
An Example
Example
Find f ′(x) for f (x) = x2(x− 4)5.
f ′(x) = x2 ddx
(x− 4)5 + (x− 4)5 ddx
x2
= x2(5(x− 4)4) + (x− 4)5(2x)
= x(x− 4)4(5x + (x− 4)(2))
= x(x− 4)4(5x + 2x− 8)
= x(x− 4)4(7x− 8)
An Example
Example
Find f ′(x) for f (x) = x2(x− 4)5.
f ′(x) = x2 ddx
(x− 4)5 + (x− 4)5 ddx
x2
= x2(5(x− 4)4) + (x− 4)5(2x)
= x(x− 4)4(5x + (x− 4)(2))
= x(x− 4)4(5x + 2x− 8)
= x(x− 4)4(7x− 8)
An Example
Example
Find f ′(x) for f (x) = x2(x− 4)5.
f ′(x) = x2 ddx
(x− 4)5 + (x− 4)5 ddx
x2
= x2(5(x− 4)4) + (x− 4)5(2x)
= x(x− 4)4(5x + (x− 4)(2))
= x(x− 4)4(5x + 2x− 8)
= x(x− 4)4(7x− 8)
The Quotient Rule
Let Q(x) = f (x)g(x) .
f (x) = Q(x)g(x)
f ′(x) = Q(x)g′(x) + Q′(x)g(x)
f ′(x) =f (x)g′(x)
g(x)+ Q′(x)g(x)
f ′(x)g(x) = f (x)g′(x) + Q′(x)[g(x)]2
Q′(x) =f ′(x)g(x)− g′(x)f (x)
[g(x)]2
The Quotient Rule
Let Q(x) = f (x)g(x) .
f (x) = Q(x)g(x)
f ′(x) = Q(x)g′(x) + Q′(x)g(x)
f ′(x) =f (x)g′(x)
g(x)+ Q′(x)g(x)
f ′(x)g(x) = f (x)g′(x) + Q′(x)[g(x)]2
Q′(x) =f ′(x)g(x)− g′(x)f (x)
[g(x)]2
The Quotient Rule
Let Q(x) = f (x)g(x) .
f (x) = Q(x)g(x)
f ′(x) = Q(x)g′(x) + Q′(x)g(x)
f ′(x) =f (x)g′(x)
g(x)+ Q′(x)g(x)
f ′(x)g(x) = f (x)g′(x) + Q′(x)[g(x)]2
Q′(x) =f ′(x)g(x)− g′(x)f (x)
[g(x)]2
The Quotient Rule
Let Q(x) = f (x)g(x) .
f (x) = Q(x)g(x)
f ′(x) = Q(x)g′(x) + Q′(x)g(x)
f ′(x) =f (x)g′(x)
g(x)+ Q′(x)g(x)
f ′(x)g(x) = f (x)g′(x) + Q′(x)[g(x)]2
Q′(x) =f ′(x)g(x)− g′(x)f (x)
[g(x)]2
The Quotient Rule
Let Q(x) = f (x)g(x) .
f (x) = Q(x)g(x)
f ′(x) = Q(x)g′(x) + Q′(x)g(x)
f ′(x) =f (x)g′(x)
g(x)+ Q′(x)g(x)
f ′(x)g(x) = f (x)g′(x) + Q′(x)[g(x)]2
Q′(x) =f ′(x)g(x)− g′(x)f (x)
[g(x)]2
The Quotient Rule
Let Q(x) = f (x)g(x) .
f (x) = Q(x)g(x)
f ′(x) = Q(x)g′(x) + Q′(x)g(x)
f ′(x) =f (x)g′(x)
g(x)+ Q′(x)g(x)
f ′(x)g(x) = f (x)g′(x) + Q′(x)[g(x)]2
Q′(x) =f ′(x)g(x)− g′(x)f (x)
[g(x)]2
The Quotient Rule
The Quotient Rule
ddx
f (x)
g(x)=
f ′(x)g(x)− g′(x)f (x)
[g(x)]2
Alternate DefinitionLet u = f (x) and v = g(x). Then
ddx
(uv
)=
vu′ − uv′
v2
The Quotient Rule
The Quotient Rule
ddx
f (x)
g(x)=
f ′(x)g(x)− g′(x)f (x)
[g(x)]2
Alternate DefinitionLet u = f (x) and v = g(x). Then
ddx
(uv
)=
vu′ − uv′
v2
Examples
Example
Find the equation of the tangent line to f (x) = x+1x at the point where
x = 3.
f ′(x) =x d
dx(x + 1)− (x + 1) ddx(x)
x2
=x(1)− (x + 1)(1)
x2
=−1x2
∣∣∣∣x=3
= −19
Examples
Example
Find the equation of the tangent line to f (x) = x+1x at the point where
x = 3.
f ′(x) =
x ddx(x + 1)− (x + 1) d
dx(x)
x2
=x(1)− (x + 1)(1)
x2
=−1x2
∣∣∣∣x=3
= −19
Examples
Example
Find the equation of the tangent line to f (x) = x+1x at the point where
x = 3.
f ′(x) =x d
dx(x + 1)− (x + 1) ddx(x)
x2
=x(1)− (x + 1)(1)
x2
=−1x2
∣∣∣∣x=3
= −19
Examples
Example
Find the equation of the tangent line to f (x) = x+1x at the point where
x = 3.
f ′(x) =x d
dx(x + 1)− (x + 1) ddx(x)
x2
=x(1)− (x + 1)(1)
x2
=−1x2
∣∣∣∣x=3
= −19
Examples
Example
Find the equation of the tangent line to f (x) = x+1x at the point where
x = 3.
f ′(x) =x d
dx(x + 1)− (x + 1) ddx(x)
x2
=x(1)− (x + 1)(1)
x2
=−1x2
∣∣∣∣x=3
= −19
Examples
Example
Find the equation of the tangent line to f (x) = x+1x at the point where
x = 3.
f ′(x) =x d
dx(x + 1)− (x + 1) ddx(x)
x2
=x(1)− (x + 1)(1)
x2
=−1x2
∣∣∣∣x=3
= −19
Examples
What else do we need?
y∣∣∣∣x=3
=
43
y = mx + b43
= −19
(3) + b
43
= −13
+ b
53
= b
And the tangent line is y = − 19 x + 5
3 .
Examples
What else do we need?
y∣∣∣∣x=3
=43
y = mx + b43
= −19
(3) + b
43
= −13
+ b
53
= b
And the tangent line is y = − 19 x + 5
3 .
Examples
What else do we need?
y∣∣∣∣x=3
=43
y = mx + b43
= −19
(3) + b
43
= −13
+ b
53
= b
And the tangent line is y = − 19 x + 5
3 .
Examples
What else do we need?
y∣∣∣∣x=3
=43
y = mx + b43
= −19
(3) + b
43
= −13
+ b
53
= b
And the tangent line is y = − 19 x + 5
3 .
Normal Lines
Example
Find the equation of the normal line to the curve f (x) = x+1x at the
point x = 3.
Anyone know what is a normal line?
DefinitionThe normal line to a curve at a point is the line perpendicular to thatcurve at the point. That is, it is the line perpendicular to the tangentline at the point of tangency.
Normal Lines
Example
Find the equation of the normal line to the curve f (x) = x+1x at the
point x = 3.
Anyone know what is a normal line?
DefinitionThe normal line to a curve at a point is the line perpendicular to thatcurve at the point. That is, it is the line perpendicular to the tangentline at the point of tangency.
Normal Lines
Example
Find the equation of the normal line to the curve f (x) = x+1x at the
point x = 3.
Anyone know what is a normal line?
DefinitionThe normal line to a curve at a point is the line perpendicular to thatcurve at the point. That is, it is the line perpendicular to the tangentline at the point of tangency.
Normal Lines
x
y
Normal Lines
x
y
Normal Lines
x
y
Normal Lines
x
y
Normal Lines
If the tangent line has a slope of −19 at the point where x = 3, what is
the slope of the normal line at the same point?
m = 9
y = mx + b43
= 9(3) + b
43
= 27 + b
− 773
= b
And so the equation of the normal line is y = 9x− 773 .
Normal Lines
If the tangent line has a slope of −19 at the point where x = 3, what is
the slope of the normal line at the same point? m = 9
y = mx + b43
= 9(3) + b
43
= 27 + b
− 773
= b
And so the equation of the normal line is y = 9x− 773 .
Normal Lines
If the tangent line has a slope of −19 at the point where x = 3, what is
the slope of the normal line at the same point? m = 9
y = mx + b
43
= 9(3) + b
43
= 27 + b
− 773
= b
And so the equation of the normal line is y = 9x− 773 .
Normal Lines
If the tangent line has a slope of −19 at the point where x = 3, what is
the slope of the normal line at the same point? m = 9
y = mx + b43
= 9(3) + b
43
= 27 + b
− 773
= b
And so the equation of the normal line is y = 9x− 773 .
Normal Lines
If the tangent line has a slope of −19 at the point where x = 3, what is
the slope of the normal line at the same point? m = 9
y = mx + b43
= 9(3) + b
43
= 27 + b
− 773
= b
And so the equation of the normal line is y = 9x− 773 .
Normal Lines
If the tangent line has a slope of −19 at the point where x = 3, what is
the slope of the normal line at the same point? m = 9
y = mx + b43
= 9(3) + b
43
= 27 + b
− 773
= b
And so the equation of the normal line is y = 9x− 773 .
Higher Order Derivatives
What do you think it means to take the 2nd derivative?
Example
Find d2
dx2 (3x4 − 2x2).
d2
dx2 (3x4 − 2x2) =ddx
(12x3 − 4x)
= 36x2 − 4
Higher Order Derivatives
What do you think it means to take the 2nd derivative?
Example
Find d2
dx2 (3x4 − 2x2).
d2
dx2 (3x4 − 2x2) =ddx
(12x3 − 4x)
= 36x2 − 4
Higher Order Derivatives
What do you think it means to take the 2nd derivative?
Example
Find d2
dx2 (3x4 − 2x2).
d2
dx2 (3x4 − 2x2) =
ddx
(12x3 − 4x)
= 36x2 − 4
Higher Order Derivatives
What do you think it means to take the 2nd derivative?
Example
Find d2
dx2 (3x4 − 2x2).
d2
dx2 (3x4 − 2x2) =ddx
(12x3 − 4x)
= 36x2 − 4
Higher Order Derivatives
What do you think it means to take the 2nd derivative?
Example
Find d2
dx2 (3x4 − 2x2).
d2
dx2 (3x4 − 2x2) =ddx
(12x3 − 4x)
= 36x2 − 4
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) = 6x
f (3)(x) = 6
f (4)(x) = 0
f (k)(x) for all k > 4 is 0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) = 6x
f (3)(x) = 6
f (4)(x) = 0
f (k)(x) for all k > 4 is 0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) =
3x2 + 3
f ′′(x) = 6x
f (3)(x) = 6
f (4)(x) = 0
f (k)(x) for all k > 4 is 0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) = 6x
f (3)(x) = 6
f (4)(x) = 0
f (k)(x) for all k > 4 is 0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) =
6x
f (3)(x) = 6
f (4)(x) = 0
f (k)(x) for all k > 4 is 0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) = 6x
f (3)(x) = 6
f (4)(x) = 0
f (k)(x) for all k > 4 is 0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) = 6x
f (3)(x) =
6
f (4)(x) = 0
f (k)(x) for all k > 4 is 0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) = 6x
f (3)(x) = 6
f (4)(x) = 0
f (k)(x) for all k > 4 is 0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) = 6x
f (3)(x) = 6
f (4)(x) =
0
f (k)(x) for all k > 4 is 0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) = 6x
f (3)(x) = 6
f (4)(x) = 0
f (k)(x) for all k > 4 is 0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) = 6x
f (3)(x) = 6
f (4)(x) = 0
f (k)(x) for all k > 4 is
0
Higher Order Derivatives
Example
Find derivatives of all orders for f (x) = x3 + 3x + 1.
f (0)(x) = x3 + 3x + 1
f ′(x) = 3x2 + 3
f ′′(x) = 6x
f (3)(x) = 6
f (4)(x) = 0
f (k)(x) for all k > 4 is 0