Chapter 3: Differentiation (p 39-45)
MATH160Lecture 13.2:
Calculus - Derivative rules examples
Sarah Wakes
University of Otago
16 September 2021
Chapter 3: Differentiation (p 39-45)
Practice
Exercise: differentiate the following functions
f (x) = cos x3
y =3t2 + 2
√t
t
q(x) = csc x ln x
h(s) =√s3(1 + s2)
y = ex2
g(x) =
√1 + x
x2 − 3
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 1
f (x) = cos(x3)
Let u = x3, then f (u) = cos(u) = y .
Chain rule: dydx = dy
dududx
dydu = − sin u, du
dx = 3x2.dydx = − sin(u) · 3x2= − sin(x3) · 3x2.
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 1
f (x) = cos(x3)
Let u = x3,
then f (u) = cos(u) = y .
Chain rule: dydx = dy
dududx
dydu = − sin u, du
dx = 3x2.dydx = − sin(u) · 3x2= − sin(x3) · 3x2.
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 1
f (x) = cos(x3)
Let u = x3, then f (u) = cos(u) = y .
Chain rule: dydx = dy
dududx
dydu = − sin u, du
dx = 3x2.dydx = − sin(u) · 3x2= − sin(x3) · 3x2.
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 1
f (x) = cos(x3)
Let u = x3, then f (u) = cos(u) = y .
Chain rule: dydx = dy
dududx
dydu = − sin u, du
dx = 3x2.dydx = − sin(u) · 3x2= − sin(x3) · 3x2.
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 1
f (x) = cos(x3)
Let u = x3, then f (u) = cos(u) = y .
Chain rule: dydx = dy
dududx
dydu = − sin u,
dudx = 3x2.
dydx = − sin(u) · 3x2= − sin(x3) · 3x2.
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 1
f (x) = cos(x3)
Let u = x3, then f (u) = cos(u) = y .
Chain rule: dydx = dy
dududx
dydu = − sin u, du
dx = 3x2.
dydx = − sin(u) · 3x2= − sin(x3) · 3x2.
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 1
f (x) = cos(x3)
Let u = x3, then f (u) = cos(u) = y .
Chain rule: dydx = dy
dududx
dydu = − sin u, du
dx = 3x2.dydx = − sin(u) · 3x2
= − sin(x3) · 3x2.
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 1
f (x) = cos(x3)
Let u = x3, then f (u) = cos(u) = y .
Chain rule: dydx = dy
dududx
dydu = − sin u, du
dx = 3x2.dydx = − sin(u) · 3x2= − sin(x3) · 3x2.
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 2
y = 3t2+2√t
t
Remember:√t = t1/2
Quotient rule:(
fg
)′(t) = f ′(t)g(t)−f (t)g ′(t)
g(t)2
f (t) = 3t2 + 2t1/2, so f ′(t) = 3 · 2t + 2 · 12 t−1/2
g(t) = t, so g ′(t) = 1.
Therefore:
dy
dt=
(f
g
)′(t) =
(6t + t−1/2)t − (3t2 + 2t1/2) · 1t2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 2
y = 3t2+2√t
t Remember:√t = t1/2
Quotient rule:(
fg
)′(t) = f ′(t)g(t)−f (t)g ′(t)
g(t)2
f (t) = 3t2 + 2t1/2, so f ′(t) = 3 · 2t + 2 · 12 t−1/2
g(t) = t, so g ′(t) = 1.
Therefore:
dy
dt=
(f
g
)′(t) =
(6t + t−1/2)t − (3t2 + 2t1/2) · 1t2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 2
y = 3t2+2√t
t Remember:√t = t1/2
Quotient rule:(
fg
)′(t) = f ′(t)g(t)−f (t)g ′(t)
g(t)2
f (t) = 3t2 + 2t1/2, so f ′(t) = 3 · 2t + 2 · 12 t−1/2
g(t) = t, so g ′(t) = 1.
Therefore:
dy
dt=
(f
g
)′(t) =
(6t + t−1/2)t − (3t2 + 2t1/2) · 1t2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 2
y = 3t2+2√t
t Remember:√t = t1/2
Quotient rule:(
fg
)′(t) = f ′(t)g(t)−f (t)g ′(t)
g(t)2
f (t) = 3t2 + 2t1/2, so
f ′(t) = 3 · 2t + 2 · 12 t−1/2
g(t) = t, so g ′(t) = 1.
Therefore:
dy
dt=
(f
g
)′(t) =
(6t + t−1/2)t − (3t2 + 2t1/2) · 1t2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 2
y = 3t2+2√t
t Remember:√t = t1/2
Quotient rule:(
fg
)′(t) = f ′(t)g(t)−f (t)g ′(t)
g(t)2
f (t) = 3t2 + 2t1/2, so f ′(t) = 3 · 2t + 2 · 12 t−1/2
g(t) = t, so g ′(t) = 1.
Therefore:
dy
dt=
(f
g
)′(t) =
(6t + t−1/2)t − (3t2 + 2t1/2) · 1t2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 2
y = 3t2+2√t
t Remember:√t = t1/2
Quotient rule:(
fg
)′(t) = f ′(t)g(t)−f (t)g ′(t)
g(t)2
f (t) = 3t2 + 2t1/2, so f ′(t) = 3 · 2t + 2 · 12 t−1/2
g(t) = t, so
g ′(t) = 1.
Therefore:
dy
dt=
(f
g
)′(t) =
(6t + t−1/2)t − (3t2 + 2t1/2) · 1t2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 2
y = 3t2+2√t
t Remember:√t = t1/2
Quotient rule:(
fg
)′(t) = f ′(t)g(t)−f (t)g ′(t)
g(t)2
f (t) = 3t2 + 2t1/2, so f ′(t) = 3 · 2t + 2 · 12 t−1/2
g(t) = t, so g ′(t) = 1.
Therefore:
dy
dt=
(f
g
)′(t)
=(6t + t−1/2)t − (3t2 + 2t1/2) · 1
t2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 2
y = 3t2+2√t
t Remember:√t = t1/2
Quotient rule:(
fg
)′(t) = f ′(t)g(t)−f (t)g ′(t)
g(t)2
f (t) = 3t2 + 2t1/2, so f ′(t) = 3 · 2t + 2 · 12 t−1/2
g(t) = t, so g ′(t) = 1.
Therefore:
dy
dt=
(f
g
)′(t) =
(6t + t−1/2)t − (3t2 + 2t1/2) · 1t2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3
q(x) = csc(x) ln x
= h(x)g(x) where h(x) = csc(x) andg(x) = ln x .
Product rule:q′(x) = h′(x)g(x) + h(x)g ′(x)
h(x) = csc(x) = 1sin x
Reciprocal rule:(
1p
)′= −p′
p2 .
In this case p(x) = sin x , p′(x) = cos x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3
q(x) = csc(x) ln x = h(x)g(x)
where h(x) = csc(x) andg(x) = ln x .
Product rule:q′(x) = h′(x)g(x) + h(x)g ′(x)
h(x) = csc(x) = 1sin x
Reciprocal rule:(
1p
)′= −p′
p2 .
In this case p(x) = sin x , p′(x) = cos x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3
q(x) = csc(x) ln x = h(x)g(x) where h(x) = csc(x) andg(x) = ln x .
Product rule:q′(x) = h′(x)g(x) + h(x)g ′(x)
h(x) = csc(x) = 1sin x
Reciprocal rule:(
1p
)′= −p′
p2 .
In this case p(x) = sin x , p′(x) = cos x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3
q(x) = csc(x) ln x = h(x)g(x) where h(x) = csc(x) andg(x) = ln x .
Product rule:q′(x) = h′(x)g(x) + h(x)g ′(x)
h(x) = csc(x) =
1sin x
Reciprocal rule:(
1p
)′= −p′
p2 .
In this case p(x) = sin x , p′(x) = cos x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3
q(x) = csc(x) ln x = h(x)g(x) where h(x) = csc(x) andg(x) = ln x .
Product rule:q′(x) = h′(x)g(x) + h(x)g ′(x)
h(x) = csc(x) = 1sin x
Reciprocal rule:(
1p
)′= −p′
p2 .
In this case p(x) = sin x , p′(x) = cos x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3
q(x) = csc(x) ln x = h(x)g(x) where h(x) = csc(x) andg(x) = ln x .
Product rule:q′(x) = h′(x)g(x) + h(x)g ′(x)
h(x) = csc(x) = 1sin x
Reciprocal rule:(
1p
)′= −p′
p2 .
In this case p(x) = sin x , p′(x) = cos x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3
q(x) = csc(x) ln x = h(x)g(x) where h(x) = csc(x) andg(x) = ln x .
Product rule:q′(x) = h′(x)g(x) + h(x)g ′(x)
h(x) = csc(x) = 1sin x
Reciprocal rule:(
1p
)′= −p′
p2 .
In this case p(x) = sin x , p′(x) = cos x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3 continuedddx
(1
sin x
)=
− cos xsin2 x
= − cos xsin x
1sin x = − cot x csc x= h′(x).
q(x) = ln xso q′(x) = 1x .
Product rule:
q′(x) = h′(x)g(x) + h(x)g ′(x)
= − cot x csc x · ln x + csc xx
= csc x(− cot x ln x + 1
x
)
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3 continuedddx
(1
sin x
)= − cos x
sin2 x
= − cos xsin x
1sin x = − cot x csc x= h′(x).
q(x) = ln xso q′(x) = 1x .
Product rule:
q′(x) = h′(x)g(x) + h(x)g ′(x)
= − cot x csc x · ln x + csc xx
= csc x(− cot x ln x + 1
x
)
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3 continuedddx
(1
sin x
)= − cos x
sin2 x= − cos x
sin x1
sin x
= − cot x csc x= h′(x).
q(x) = ln xso q′(x) = 1x .
Product rule:
q′(x) = h′(x)g(x) + h(x)g ′(x)
= − cot x csc x · ln x + csc xx
= csc x(− cot x ln x + 1
x
)
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3 continuedddx
(1
sin x
)= − cos x
sin2 x= − cos x
sin x1
sin x = − cot x csc x
= h′(x).
q(x) = ln xso q′(x) = 1x .
Product rule:
q′(x) = h′(x)g(x) + h(x)g ′(x)
= − cot x csc x · ln x + csc xx
= csc x(− cot x ln x + 1
x
)
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3 continuedddx
(1
sin x
)= − cos x
sin2 x= − cos x
sin x1
sin x = − cot x csc x= h′(x).
q(x) = ln x
so q′(x) = 1x .
Product rule:
q′(x) = h′(x)g(x) + h(x)g ′(x)
= − cot x csc x · ln x + csc xx
= csc x(− cot x ln x + 1
x
)
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3 continuedddx
(1
sin x
)= − cos x
sin2 x= − cos x
sin x1
sin x = − cot x csc x= h′(x).
q(x) = ln xso q′(x) = 1x .
Product rule:
q′(x) = h′(x)g(x) + h(x)g ′(x)
= − cot x csc x · ln x + csc xx
= csc x(− cot x ln x + 1
x
)
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3 continuedddx
(1
sin x
)= − cos x
sin2 x= − cos x
sin x1
sin x = − cot x csc x= h′(x).
q(x) = ln xso q′(x) = 1x .
Product rule:
q′(x) = h′(x)g(x) + h(x)g ′(x)
= − cot x csc x · ln x + csc xx
= csc x(− cot x ln x + 1
x
)
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3 continuedddx
(1
sin x
)= − cos x
sin2 x= − cos x
sin x1
sin x = − cot x csc x= h′(x).
q(x) = ln xso q′(x) = 1x .
Product rule:
q′(x) = h′(x)g(x) + h(x)g ′(x)
= − cot x csc x · ln x +
csc xx
= csc x(− cot x ln x + 1
x
)
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3 continuedddx
(1
sin x
)= − cos x
sin2 x= − cos x
sin x1
sin x = − cot x csc x= h′(x).
q(x) = ln xso q′(x) = 1x .
Product rule:
q′(x) = h′(x)g(x) + h(x)g ′(x)
= − cot x csc x · ln x + csc xx
= csc x(− cot x ln x + 1
x
)
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 3 continuedddx
(1
sin x
)= − cos x
sin2 x= − cos x
sin x1
sin x = − cot x csc x= h′(x).
q(x) = ln xso q′(x) = 1x .
Product rule:
q′(x) = h′(x)g(x) + h(x)g ′(x)
= − cot x csc x · ln x + csc xx
= csc x(− cot x ln x + 1
x
)
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 4
y = ex2.
Let u = x2. Then y = eu. Chain rule:
dydx = dy
dududx
= eu · 2x
= ex2 · 2x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 4
y = ex2.
Let u = x2. Then y = eu.
Chain rule:
dydx = dy
dududx
= eu · 2x
= ex2 · 2x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 4
y = ex2.
Let u = x2. Then y = eu. Chain rule:
dydx = dy
dududx
= eu · 2x
= ex2 · 2x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 4
y = ex2.
Let u = x2. Then y = eu. Chain rule:
dydx = dy
dududx
= eu ·
2x
= ex2 · 2x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 4
y = ex2.
Let u = x2. Then y = eu. Chain rule:
dydx = dy
dududx
= eu · 2x
= ex2 · 2x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 4
y = ex2.
Let u = x2. Then y = eu. Chain rule:
dydx = dy
dududx
= eu · 2x
= ex2 · 2x .
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then g(x) =√u = u1/2.
dgdu = 1
2u−1/2= 1
2√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then
g(x) =√u = u1/2.
dgdu = 1
2u−1/2= 1
2√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then g(x) =√u =
u1/2.
dgdu = 1
2u−1/2= 1
2√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then g(x) =√u = u1/2.
dgdu = 1
2u−1/2= 1
2√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then g(x) =√u = u1/2.
dgdu =
12u−1/2= 1
2√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then g(x) =√u = u1/2.
dgdu = 1
2u−1/2
= 12√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then g(x) =√u = u1/2.
dgdu = 1
2u−1/2= 1
2√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then g(x) =√u = u1/2.
dgdu = 1
2u−1/2= 1
2√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then g(x) =√u = u1/2.
dgdu = 1
2u−1/2= 1
2√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then g(x) =√u = u1/2.
dgdu = 1
2u−1/2= 1
2√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5
g(x) =√
1+xx2−3
.
Chain rule:
Let u = 1+xx2−3
, then g(x) =√u = u1/2.
dgdu = 1
2u−1/2= 1
2√u
.
dudx =
ddx
(1+x)·(x2−3)−(1+x)· ddx
(x2−3)
(x2−3)2
= 1·(x2−3)−(1+x)·2x(x2−3)2
= x2−3−2x−2x2
(x2−3)2
= −x2−2x−3(x2−3)2
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5 continued
Chain rule:
dg
dx=
dg
du
du
dx
=1
2√
1+xx2−3
[−x2 − 2x − 3
(x2 − 3)2
]
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5 continued
Chain rule:
dg
dx=
dg
du
du
dx
=1
2√
1+xx2−3
[−x2 − 2x − 3
(x2 − 3)2
]
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5 continued
Chain rule:
dg
dx=
dg
du
du
dx
=1
2√
1+xx2−3
[−x2 − 2x − 3
(x2 − 3)2
]
MATH160 - Lecture 13.2 90
Chapter 3: Differentiation (p 39-45)
Practice
Exercise 5 continued
Chain rule:
dg
dx=
dg
du
du
dx
=1
2√
1+xx2−3
[−x2 − 2x − 3
(x2 − 3)2
]
MATH160 - Lecture 13.2 90