MATH 31
UNIT 3 LESSON #4
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Part 2: Tangent, Cotangent, Secant, and Cosecant
NAME ANSWERS
Page 1 of 5
The derivative of f(x) = tan x can be found by first rewriting the function in terms of
sine and cosine.
f  x   tan x 
sin x
cos x
Now, using the Quotient Rule:
cos x  cos x   sin x   sin x 
f ' x 
cos 2 x
cos 2 x  sin 2 x
f ' x 
cos 2 x
Using the Pythagorean Identity:
1
f ' x 
 sec 2 x
2
cos x
The derivative of f(x) = cot x can be found by first rewriting the function in terms of
sine and cosine.
f  x   cot x 
cos x
sin x
Now, using the Quotient Rule:
 sin x  sin x   cos x  cos x 
f ' x 
sin 2 x
 sin 2 x  cos 2 x
f ' x 
sin 2 x
Using the Pythagorean Identity:
1
f ' x 
 csc 2 x
2
sin x
U3 L4 Derivatives of Tan & Recip Fn
MATH 31
UNIT 3 LESSON #4
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Part 2: Tangent, Cotangent, Secant, and Cosecant
NAME ANSWERS
Page 2 of 5
The derivative of the secant and cosecant functions can also be found by first rewriting
the function in terms of sine and cosine.
Find the derivative of
f  x   sec x 
1
1
  cos x 
cos x
Now, using the Chain Rule:
f '  x     cos x 
2
  sin x 
Find the derivative of
f  x   csc x 
1
1
  sin x 
sin x
Now, using the Chain Rule:
f '  x   1 sin x 
2
 cos x 
f ' x 
sin x
cos 2 x
f ' x 
 cos x
sin 2 x
f ' x 
sin x
1

cos x cos x
f ' x 
 cos x
1

sin x sin x
f '  x   tan x sec x
f '  x    cot x csc x
EXAMPLE 1: Find the derivative of the function y = 2sec (3x).
dy
 2 tan  3 x  sec  3 x  3
dx
dy
 6 tan  3 x  sec  3 x 
dx
U3 L4 Derivatives of Tan & Recip Fn
MATH 31
UNIT 3 LESSON #4
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Part 2: Tangent, Cotangent, Secant, and Cosecant
NAME ANSWERS
Page 3 of 5
EXAMPLE 2: Find the derivative of the function f(x) = sin (2x) tan x
Use the Product Rule and the Chain Rule
g  x   sin  2 x 
g '  x   cos  2 x  2
h  x   tan x
h '  x   sec2 x




f '  x   sin  2 x  sec 2 x  tan x  2 cos  2 x  
f '  x   sin  2 x  sec 2 x  2 tan x cos  2 x 
EXAMPLE 3: Find the derivative of the function
Use the Quotient Rule and the Chain Rule:
U3 L4 Derivatives of Tan & Recip Fn
g '  x   2cos  2 x 
MATH 31
UNIT 3 LESSON #4
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Part 2: Tangent, Cotangent, Secant, and Cosecant
NAME ANSWERS
Page 4 of 5
Assignment Questions
1. Differentiate each of the following trigonometric functions.
a) y = 2cot(2x)
dy
 4 csc 2 (2 x)
dx
b) f(x) = sin x sec (2x)
f '( x)  sin x sec(2 x) tan(2 x)(2)  sec(2 x) cos x
f '( x)  sec(2 x)(2sin x tan 2 x  cos x)
c) y 
tan 5 x  4 
cos x 3
 
3
2
3
2
dy cos x sec  5 x  4  (5)  tan(5 x  4)[ sin( x )(3x )]

dx
cos 2 x3
 
3
2
2
3
dy 5cos x sec  5 x  4   3x tan(5 x  4)sin( x )

dx
cos2 x3
 
d) f(x) = csc3(3x2)  csc3x 2 
3
 
    6x 
f '( x)  3csc2 3x 2  csc 3x 2 cot 3x 2
   
f '( x)  18 x csc3 3 x 2 cot 3x 2
U3 L4 Derivatives of Tan & Recip Fn
MATH 31
UNIT 3 LESSON #4
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Part 2: Tangent, Cotangent, Secant, and Cosecant
NAME ANSWERS
Page 5 of 5

2. Find the slope of the tangent to the function f x  tan 3x when x  .
3
2
f '  x   3sec  3x 
 
f '  x   3sec 2  3    3sec 2 
3

2
 
f '    3  1  3
3

 x
3. Find the slope of the tangent to the function y  sin x tan   when x  .
3
2
dy
x
 x  1 
 tan cos x  sin x sec 2   
dx
2
 2  2 
dy
x
1
x
 tan cos x  sin x sec 2  
dx
2
2
2

 1 
 
 tan cos  sin sec 2  
6
3 2
3
6
 1  1  1  3  2 


   

 3   2  2  2   3 
1
1

2 3
3
1
2


2 3 2 3
3

 0.866
2 3
4. Find the slope of the tangent line to the function y  csc 2 x when x  
dy
  cot(2 x) csc(2 x)(2)
dx
 
 
2 cot    csc   
 4
 4


2  1  2  2 2  2.83
U3 L4 Derivatives of Tan & Recip Fn

8
.
2