Part 2

3.5: Derivatives of Trigonometric Functions

Part 2: The Other Basic Functions

MATH 165: Calculus I

Department of Mathematics

Iowa State University

Paul J. Barloon

MATH 165 Section 3.5

Part 2 The Other Basic Functions

A Brief Review

Recall the derivatives of sin(x) and cos(x):

d

dx

[sin(x)] = cos(x)

d

dx

[cos(x)] = � sin(x)

Using these and our di↵erentiation rules, we get the derivatives ofthe remaining trigonometric functions . . .

MATH 165 Section 3.5

Part 2 The Other Basic Functions

A Brief Review

Recall the derivatives of sin(x) and cos(x):

d

dx

[sin(x)] = cos(x)

d

dx

[cos(x)] = � sin(x)

Using these and our di↵erentiation rules, we get the derivatives ofthe remaining trigonometric functions . . .

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Derivative of tan(x)

For f (x) = tan(x), use the definition of the tangent function andthe Quotient Rule:

d

dx

[tan(x)] =d

dx

sin(x)

cos(x)

�

=cos(x)[cos(x)]� sin(x)[� sin(x)]

cos2(x)

=cos2(x) + sin2(x)

cos2(x)

=1

cos2(x)= sec2(x)

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Derivative of tan(x)

For f (x) = tan(x), use the definition of the tangent function andthe Quotient Rule:

d

dx

[tan(x)] =d

dx

sin(x)

cos(x)

�

=cos(x)[cos(x)]� sin(x)[� sin(x)]

cos2(x)

=cos2(x) + sin2(x)

cos2(x)

=1

cos2(x)= sec2(x)

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Derivative of tan(x)

For f (x) = tan(x), use the definition of the tangent function andthe Quotient Rule:

d

dx

[tan(x)] =d

dx

sin(x)

cos(x)

�

=cos(x)[cos(x)]� sin(x)[� sin(x)]

cos2(x)

=cos2(x) + sin2(x)

cos2(x)

=1

cos2(x)= sec2(x)

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Derivative of tan(x)

For f (x) = tan(x), use the definition of the tangent function andthe Quotient Rule:

d

dx

[tan(x)] =d

dx

sin(x)

cos(x)

�

=cos(x)[cos(x)]� sin(x)[� sin(x)]

cos2(x)

=cos2(x) + sin2(x)

cos2(x)

=1

cos2(x)= sec2(x)

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Derivative of sec(x)

Follow a similar procedure to find the derivative of f (x) = sec(x):

d

dx

[sec(x)] =d

dx

1

cos(x)

�

=cos(x)[0]� (1)[� sin(x)]

cos2(x)

=sin(x)

cos2(x)

=1

cos(x)· sin(x)cos(x)

= sec(x) tan(x)

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Derivative of sec(x)

Follow a similar procedure to find the derivative of f (x) = sec(x):

d

dx

[sec(x)] =d

dx

1

cos(x)

�

=cos(x)[0]� (1)[� sin(x)]

cos2(x)

=sin(x)

cos2(x)

=1

cos(x)· sin(x)cos(x)

= sec(x) tan(x)

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Derivative of sec(x)

Follow a similar procedure to find the derivative of f (x) = sec(x):

d

dx

[sec(x)] =d

dx

1

cos(x)

�

=cos(x)[0]� (1)[� sin(x)]

cos2(x)

=sin(x)

cos2(x)

=1

cos(x)· sin(x)cos(x)

= sec(x) tan(x)

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Derivative of sec(x)

Follow a similar procedure to find the derivative of f (x) = sec(x):

d

dx

[sec(x)] =d

dx

1

cos(x)

�

=cos(x)[0]� (1)[� sin(x)]

cos2(x)

=sin(x)

cos2(x)

=1

cos(x)· sin(x)cos(x)

= sec(x) tan(x)

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Derivative of cot(x) and csc(x)

Use the same techniques to get the final two trigonometricderivatives:

d

dx

[cot(x)] = � csc2(x)

d

dx

[csc(x)] = � csc(x) cot(x)

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Summary

Here is the complete list.

d

dx

[sin(x)] = cos(x)

d

dx

[tan(x)] = sec2(x)

d

dx

[sec(x)] = sec(x) tan(x)

d

dx

[cos(x)] = � sin(x)

d

dx

[cot(x)] = � csc2(x)

d

dx

[csc(x)] = � csc(x) cot(x)

MATH 165 Section 3.5

Part 2 The Other Basic Functions

EXAMPLE 1: Finddr

d✓if r = (4 + sec ✓) sin ✓.

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Quiz Yourself

Finddp

dq

if p =sin q + cos q

sin q

A)cos q � sin q

cos q

B)� sin q � cos q

sin2 q

C) sec2 q

D) � csc2 q

E) �1

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Some Good News!

Since all of the trigonometric functions are di↵erentiable, they arecontinuous – at all points where they are defined.

Remember that finding limits of continuous functions – andcomposites of continuous functions – boils down to “plugging in”the value.

So, limits involving trigonometric functions are generallystraightforward . . .

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Some Good News!

Since all of the trigonometric functions are di↵erentiable, they arecontinuous – at all points where they are defined.

Remember that finding limits of continuous functions – andcomposites of continuous functions – boils down to “plugging in”the value.

So, limits involving trigonometric functions are generallystraightforward . . .

MATH 165 Section 3.5

Part 2 The Other Basic Functions

Some Good News!

Since all of the trigonometric functions are di↵erentiable, they arecontinuous – at all points where they are defined.

Remember that finding limits of continuous functions – andcomposites of continuous functions – boils down to “plugging in”the value.

So, limits involving trigonometric functions are generallystraightforward . . .

MATH 165 Section 3.5

Part 2 The Other Basic Functions

EXAMPLE 2: Evaluate the following limit:

limx!0

seche

x + ⇡ tan⇣ ⇡

4 sec x

⌘� 1

i

MATH 165 Section 3.5

Part 2 The Other Basic Functions

The End

MATH 165 Section 3.5

Di↵erentiation

Practice With DerivativesPractice Problems

MATH 165: Calculus I

Department of Mathematics

Iowa State University

Paul J. Barloon

MATH 165

Di↵erentiation Practice Problems

Problem 1: Suppose that a spill is forming acircular oil slick around a leaking tanker.

How fast does the area of the slick change withrespect to its radius when that radius is 1 foot?

A) 1 ft2/ft

B) ⇡ ft2/ft

C) 2⇡ ft2/ft

D) 90⇡ ft2/ft

E) 180⇡ ft2/ft

MATH 165

Di↵erentiation Practice Problems

Problem 1: Suppose that a spill is forming acircular oil slick around a leaking tanker.

How fast does the area of the slick change withrespect to its radius when that radius is 1 foot?

A) 1 ft2/ft

B) ⇡ ft2/ft

*C) 2⇡ ft2/ft

D) 90⇡ ft2/ft

E) 180⇡ ft2/ft

MATH 165

Di↵erentiation Practice Problems

Problem 2: Let f (x) =x

4 � x

3 + 5x2 � 3x + 6

x

2 � x + 2.

Find f

0(x).

A) f

0(x) =2x5 � x

4 + 3x3 � 7x + 5

(x2 � x + 2)2

B) f

0(x) =4x3 � 3x2 + 10x � 3

2x � 1

C) f

0(x) = x

2 + 3

D) f

0(x) = 2x

E) f

0(x) = 0

F) f

0(x) DNE

MATH 165

Di↵erentiation Practice Problems

Problem 2: Let f (x) =x

4 � x

3 + 5x2 � 3x + 6

x

2 � x + 2.

Find f

0(x).

A) f

0(x) =2x5 � x

4 + 3x3 � 7x + 5

(x2 � x + 2)2

B) f

0(x) =4x3 � 3x2 + 10x � 3

2x � 1

C) f

0(x) = x

2 + 3

*D) f

0(x) = 2x

E) f

0(x) = 0

F) f

0(x) DNE

MATH 165

Di↵erentiation Practice Problems

Problem 3: Suppose that y = x

2g(x) for some

di↵erentiable function g(x).

Find y

0.

A) y

0 = 2x g(x)

B) y

0 = x

2g

0(x)

C) y

0 = 2x g 0(x)

D) y

0 = 2x + g

0(x)

E) y

0 = x (x g(x) + 2 g 0(x))

F) y

0 = x (x g 0(x) + 2 g(x))

MATH 165

Di↵erentiation Practice Problems

Problem 3: Suppose that y = x

2g(x) for some

di↵erentiable function g(x).

Find y

0.

A) y

0 = 2x g(x)

B) y

0 = x

2g

0(x)

C) y

0 = 2x g 0(x)

D) y

0 = 2x + g

0(x)

E) y

0 = x (x g(x) + 2 g 0(x))

*F) y

0 = x (x g 0(x) + 2 g(x))

MATH 165

Di↵erentiation Practice Problems

Problem 4: Suppose that y =g(x)

x

2for some

di↵erentiable function g(x).

Find y

0.

A) y

0 =g

0(x)

2xD) y

0 =x g

0(x)� 2 g(x)

x

B) y

0 =2x

g

0(x)E) y

0 =2 g(x)� x g

0(x)

x

3

C) y

0 =2 g(x)� x g

0(x)

x

F) y

0 =x g

0(x)� 2 g(x)

x

3

MATH 165

Di↵erentiation Practice Problems

Problem 4: Suppose that y =g(x)

x

2for some

di↵erentiable function g(x).

Find y

0.

A) y

0 =g

0(x)

2xD) y

0 =x g

0(x)� 2 g(x)

x

B) y

0 =2x

g

0(x)E) y

0 =2 g(x)� x g

0(x)

x

3

C) y

0 =2 g(x)� x g

0(x)

x

*F) y

0 =x g

0(x)� 2 g(x)

x

3

MATH 165

Di↵erentiation Practice Problems

Problem 5a: Letf (x) = x

20 = 10x17�⇡5x

12+ 23x

9�11.1x7+3x2�5.

Find f

(21)(x).

A) 20x19 � 170x16 � 60⇡4x

11 + 6x8 � 77.7x6 + 6x

B) 20x19 � 170x16 � 12⇡5x

11 + 6x8 � 77.7x6 + 6x

C) 20x � 10

D) 20x

E) (20!)x�1

F) 0

MATH 165

Di↵erentiation Practice Problems

Problem 5a: Letf (x) = x

20 = 10x17�⇡5x

12+ 23x

9�11.1x7+3x2�5.

Find f

(21)(x).

A) 20x19 � 170x16 � 60⇡4x

11 + 6x8 � 77.7x6 + 6x

B) 20x19 � 170x16 � 12⇡5x

11 + 6x8 � 77.7x6 + 6x

C) 20x � 10

D) 20x

E) (20!)x�1

*F) 0

MATH 165

Di↵erentiation Practice Problems

Problem 5b: Letf (x) = x

20 = 10x17�⇡5x

12+ 23x

9�11.1x7+3x2�5.

Find f

(20)(x).

A) 20x

B) (20!)x

C) 20x�1

D) 20!

E) 20

F) 0

MATH 165

Di↵erentiation Practice Problems

Problem 5b: Letf (x) = x

20 = 10x17�⇡5x

12+ 23x

9�11.1x7+3x2�5.

Find f

(20)(x).

A) 20x

B) (20!)x

C) 20x�1

*D) 20!

E) 20

F) 0

MATH 165

Di↵erentiation Practice Problems

The End

MATH 165

MATH 165 44–49 Warm-Up Question – Sep. 19, 2018

Use the Quotient Rule to find the derivative

of the following function:

f (x) =

1

cos(x)

A) f

0(x) = 0 D) f

0(x) = sec(x) tan(x)

B) f

0(x) = � 1

sin(x)

E) f

0(x) = � sin(x)

cos

2(x)

C) f

0(x) =

sin(x)

cos

2(x)

F) f

0(x) = � sec(x) tan(x)

MATH 165 44–49 Warm-Up Question – Sep. 19, 2018

Use the Quotient Rule to find the derivative

of the following function:

f (x) =

1

cos(x)

A) f

0(x) = 0 *D) f

0(x) = sec(x) tan(x)

B) f

0(x) = � 1

sin(x)

E) f

0(x) = � sin(x)

cos

2(x)

*C) f

0(x) =

sin(x)

cos

2(x)

F) f

0(x) = � sec(x) tan(x)