Calculus 1 Lecture Notes
Section 2.6
Page 1 of 6
Section 2.6: Derivatives of Trigonometric Functions
Big Idea: The shortcut for taking the derivatives of the six trigonometric functions is to memorize the
formulas in the table below.
Big Skill: You should be able to take the derivative of functions made up of combinations of
trigonometric and power functions.
d
sin x  cos x
dx
d
cot x   csc 2 x
dx
d
cos x   sin x
dx
d
sec x  sec x tan x
dx
d
tan x  sec 2 x
dx
d
csc x   csc x cot x
dx
Before we can prove these derivatives, we need to establish some helpful limits, because we stumble
upon them in applying the definition of the derivative to the sine function:
sin  x  h   sin  x 
d
sin x  lim
h 0
dx
h
sin  x  cos  h   sin  h  cos  x   sin  x 
 lim
h 0
h
sin  x  cos  h   sin  x 
sin  h  cos  x 
 lim
 lim
h 0
h 0
h
h
cos  h   1
sin  h 
 sin  x  lim
 cos  x  lim
h 0
h

0
h
h
Prove: lim sin   0 (Lemma 6.1)
 0
Notice from the figure that 0 ≤ sin θ ≤ θ.
lim 0  lim sin   lim 
 0
 0
 0
0  lim sin   0
 0
 lim sin   0
 0
(by the Squeeze Theorem)
Calculus 1 Lecture Notes
Section 2.6
Page 2 of 6
Prove: lim cos   1 (Lemma 6.2)
 0
lim cos   lim 1  sin 2 
 0
 0

 lim 1  sin 2 
 0

 1  lim sin 
 0


2
1
Prove: lim
sin 
 0
Notice from the figure that:
Area ΔOPR < Area sector OPR < Area ΔOQR
Area ΔOPR = ½(base)(height) = ½(1)(sin θ)
1
1

2
Area sector OPR =  r 2   1 
2
2
2
Area ΔOQR = ½(base)(height) = ½(1)(tan θ)
 ½ sin θ < ½ θ < ½ tan θ
Divide through by ½ sin θ
1<
1>

sin 
sin 

lim 1  lim
 0
tan 
sin 
> cos θ
sin 
 0
1  lim
<
sin 

1
 lim cos 
 0

sin 
 lim
1
 0 
(by the Squeeze Theorem)
 0

 1 (Lemma 6.3)
Calculus 1 Lecture Notes
Section 2.6
Prove: lim
 0
lim
 0
1  cos 

1  cos 

Page 3 of 6
 0 (Lemma 6.4)
 1  cos  1  cos  
 lim 


 0
1  cos  
 
 1  cos 2  
 lim 

  0  1  cos   


 sin 2 

 lim 

  0  1  cos   


sin   
sin  

  lim
lim



  0     0 1  cos  
0
Now we can complete our computation for the derivative of the sine function.
Theorem 6.1:
Proof:
d
sin x  cos x
dx
sin  x  h   sin  x 
d
sin x  lim
h 0
dx
h
sin  x  cos  h   sin  h  cos  x   sin  x 
 lim
h 0
h
sin  x  cos  h   sin  x 
sin  h  cos  x 
 lim
 lim
h 0
h 0
h
h
cos  h   1
sin  h 
 sin  x  lim
 cos  x  lim
h 0
h 0
h
h
 sin  x   0  cos  x  1
 cos x
Calculus 1 Lecture Notes
Section 2.6
Theorem 6.2:
Proof:
Page 4 of 6
d
cos x   sin x
dx
cos  x  h   cos  x 
d
cos x  lim
h 0
dx
h
cos  x  cos  h   sin  x  sin  h   cos  x 
 lim
h 0
h
cos  x  cos  h   cos  x 
sin  x  sin  h 
 lim
 lim
h 0
h 0
h
h
cos  h   1
sin  h 
 cos  x  lim
 sin  x  lim
h 0
h 0
h
h
 cos  x   0  sin  x  1
  sin x
Theorem 6.3:
d
tan x  sec 2 x
dx
Proof:
d
d  sin x 
tan x  
dx
dx  cos x 
d

d

 sin x   cos x    sin x   cos x 
dx

 dx


2
 cos x 


cos 2 x  sin 2 x
 cos x 
2
1
 cos x 
2
 sec 2 x
d
cot x   csc 2 x
dx
d
sec x  sec x tan x
dx
d
csc x   csc x cot x
dx
Calculus 1 Lecture Notes
Practice:
d
5sin  x   
1.
dx 
2.
d 2
 x  2 cos  x   
dx 
3.
d
 4 tan  x   3cot  x   
dx 
4.
d 
4 x sin  x   
dx 
d  x2 
5.

dx  csc x 
Section 2.6
Page 5 of 6
Calculus 1 Lecture Notes
6.
d
cos  2 x   
dx 
7.
d
 cos 2  x   
dx
8.
d
 cos  x 2   

dx 
Section 2.6
9. Find the equation of the tangent line to y = f(x) = cos x at x =  / 2.
10. Find f 
37 
 x
and f 
240 
 x  for
f(x) = cos x.
Page 6 of 6