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