Chapter 2
THE DERIVATIVE
.
2.5 DERIVATVES OF TRIGONOMETRIC
FUNCTIONS
• Remember theorem 1.6.5
Differentiate f(x) = sin 𝑥
𝒇 𝒙+𝒉 −𝒇(𝒙)
𝒉
𝒉→𝟎
f’(x) =𝐥𝐢𝐦
sin( 𝑥+ℎ)−sin 𝑥
𝒉
𝒉→𝟎
= 𝐥𝐢𝐦
sin 𝑥cos 𝒉+𝒄𝒐𝒔𝒙𝒔𝒊𝒏𝒉−sin 𝑥
=𝐥𝐢𝐦
𝒉
𝒉→𝟎
𝑐𝑜𝑠ℎ−1
𝑠𝑖𝑛ℎ
=𝐥𝐢𝐦[sinx(
) + cosx(
)]
ℎ
ℎ
𝒉→𝟎
𝑠𝑖𝑛ℎ
1−𝑐𝑜𝑠ℎ
=𝐥𝐢𝐦[ cosx(
) - sinx(
)]
ℎ
ℎ
𝒉→𝟎
𝑠𝑖𝑛ℎ
1−𝑐𝑜𝑠ℎ
= 𝐥𝐢𝐦cosx . 𝐥𝐢𝐦 (
) - 𝐥𝐢𝐦sinx . 𝐥𝐢𝐦 (
)
ℎ
ℎ
𝒉→𝟎
𝒉→𝟎
𝒉→𝟎
𝒉→𝟎
= (𝐥𝐢𝐦cosx)(1) – (𝐥𝐢𝐦sinx)(0)
𝒉→𝟎
= cosx
𝒉→𝟎
Formula (3)
Formula (4)
• Find the derivative
of y=3x+sin(x)−4cos(x)
• Solution
• Find the derivative of
y=3x+sin(x)−4cos(x)
• Solution
y’=3+cos(x)+4sin(x)
Example 1
Find dy/dx of y = xsinx
Solution 1. obtain
u= x ; u’ = 1
v= sinx ; v’ =cosx
𝑑𝑦
𝑑𝑥
= u’v + uv’ = sinx + xcosx
Solution 2
𝑑𝑦
𝑑𝑥
=
=
𝑑
[xsinx]
𝑑𝑥
𝑑
x [sinx] +
𝑑𝑥
sinx
= xcosx + sinx
𝑑
𝑑𝑥
[x]
Example 2
Find
𝒅𝒚
𝒅𝒙
if y
𝐬𝐢𝐧 𝒙
=
𝟏+𝐜𝐨𝐬 𝒙
Solution. Using quotient rule together with Formula (3) and
(4) we obtain
𝒅𝒚
𝒅𝒙
=
𝟏+𝐜𝐨𝐬 𝒙 .
𝒅
𝒅𝒙
[𝐬𝐢𝐧 𝒙]−𝐬𝐢𝐧 𝒙 .
𝒅𝒚
𝒅𝒙
[𝟏+𝐜𝐨𝐬 𝒙]
(𝟏+𝐜𝐨𝐬 𝒙)²
=
𝟏+𝐜𝐨𝐬 𝒙 (𝐜𝐨𝐬 𝒙)−𝐬𝐢𝐧 𝒙(− 𝐬𝐢𝐧 𝒙)
(𝟏+𝐜𝐨𝐬 𝒙)²
=
𝐜𝐨𝐬 𝒙 + 𝐜𝐨𝐬² 𝒙 + 𝐬𝐢𝐧² 𝒙
(𝟏+𝐜𝐨𝐬 𝒙)²
=
𝐜𝐨𝐬 𝒙 +𝟏
(𝟏+𝐜𝐨𝐬 𝒙)²
=
𝟏
(𝟏+𝐜𝐨𝐬 𝒙)
Formula (5) (p. 170)
Formula (6)
Formula (7)
Formula (8)
𝒅
[tan 𝑥] =
𝒅𝒙
=
=
=
𝒅
𝒅𝒙
[
sin 𝑥
cos 𝑥
cos 𝑥
𝒅
𝒅𝒙
]
[sin 𝑥] −sin 𝑥
𝒅
𝒅𝒙
[cos 𝑥]
cos² 𝑥
cos 𝑥 cos 𝑥 − sin 𝑥(− sin 𝑥)
cos² 𝑥
cos² 𝑥+sin² 𝑥
cos² 𝑥
=
1
cos² 𝑥
= sec² 𝑥
Example 3
• If f(x) = sec 𝑥 Find f’’(x)
• f’(x) = sec 𝑥 tan 𝑥
U = secx ; u’ = secx.tanx
V=tanx ; v’ = sec² 𝑥
𝑑𝑦
𝑑𝑥
= 𝑢 ′v + uv’ = (secx.tanx)tanx + secx(sec² 𝑥) OR
• f’’(x) = sec 𝑥 .
•
•
𝒅
𝒅𝒙
[tan 𝑥] + tan 𝑥 .
𝒅
[sec 𝑥]
𝒅𝒙
= sec 𝑥 . sec² 𝑥 + tan 𝑥 sec 𝑥 tan 𝑥
= sec ³𝑥 + sec 𝑥 tan² 𝑥
2.6 THE CHAIN RULE
Example 1
• Find
𝒅𝒚
𝒅𝒙
of y = cos( 𝒙³)
• Solution.
Let u=x³ and express y as y=cos 𝒖
Applying Formula (1) yields
𝒅𝒚
𝒅𝒙
=
=
𝒅𝒚 𝒅𝒖
𝒅𝒖 𝒅𝒙
𝒅
𝒅𝒖
[cos 𝑢] .
𝒅
𝒅𝒙
[x³]
= (-sin 𝑢 ) . (3x²) = -3x²sin( 𝑥³)
Example 2
• Find
𝒅𝒘
𝒅𝒕
if w = tan 𝑥 and x = 4t³ + t. ; w=tan(4t³ + t)
• Solution. In this case the chain rule computations take
the form
𝒅𝒘
𝒅𝒕
=
=
𝒅𝒘 𝒅𝒙
𝒅𝒙 𝒅𝒕
𝒅
𝒅
𝒅𝒙
𝒅𝒕
[tan 𝑥] .
[4t³ + t]
= (sec² 𝑥) . (12t² + 1)
= [sec²(4t³ + t)] . (12t² + 1)
= (12t² + 1) sec²(4t³ + t)
Example 3
a) y = tan² 𝑥
u= tanx ; u’ = sec² 𝑥
Y= 𝒖𝟐
dy/dx = 2 u u’ = 2tanx sec² 𝑥
𝒅
𝒅𝒙
[tan² 𝑥] =
b) y =
𝒅
𝒅𝒙
𝑥² + 1 = (𝑥² + 1)
𝟏
𝟐
𝟏
−𝟏
𝟐
dy/dx = 𝒖
𝒅
[ 𝑥² + 1 ] =
𝒅𝒙
[(tan 𝑥)²] = 2tan 𝑥 sec² 𝑥
u’ =
𝟏
𝟐
; let u= 𝑥² + 1; u’ = 2x
𝟏 −𝟏
𝟏
𝒖 𝟐 u’= (𝑥 2
𝟐
𝟐
1
2 𝑥²+1
· 2x =
𝑥
𝑥²+1
+ 1)
𝟏
−𝟐
(2x) =
𝑥
𝑥²+1
OR
y  5u  2
where u  3t
y  5  3t   2 y  5u  2
u  3t
y  15t  2
then y  5  3t   2
dy
 15
dt
dy
5
du
du
3
dt
15  5  3
dy dy du


dt du dt

y  9x  6x  1
2
y   3x  1
y  9x2  6x  1
y  u2
u  3x  1
dy
 18 x  6
dx
dy
 2u
du
du
3
dx
2
If u  3x  1
then y  u 2
This pattern is called the
chain rule.
dy
 2  3x  1
du
dy
 6x  2
du
18 x  6   6 x  2   3
dy dy du


dx du dx

Chain Rule:
dy dy du


dx du dx
f  x   sin x g  x   x  4
2
Find: (fog)’ at x= 2
f   x   cos x
g  2  4  4  0
g  x   2x
f   0  g  2
cos  0    2  2 
1 4  4

We could also do it this way:
f  g  x    sin  x 2  4 
y  sin  x  4 
2
y  sin u
dy
 cos u
du
u  x2  4
du
 2x
dx
dy dy du


dx du dx
dy
 cos u  2 x
dx
dy
 cos  x 2  4   2 x
dx
dy
 cos  22  4   2  2
dx
dy
 cos  0   4
dx
dy
4
dx

Here is a faster way to find the derivative:
y  sin  x 2  4 
d 2
y  cos  x  4    x  4 
dx
2
y  cos  x 2  4   2 x
Differentiate the outside function...
…then the inside function
At x  2, y  4

Another example:
d
cos 2  3x 
dx
2
d
cos  3 x  
dx
It looks like we need to use the
chain rule again!
d
2 cos  3x    cos  3 x 
dx
derivative of the
outside function
derivative of the
inside function

Another example:
d
cos 2  3x 
dx
2
d
cos  3 x  
dx
d
2 cos  3x    cos  3 x 
dx
d
2 cos  3x    sin  3 x    3 x 
dx
2cos  3x   sin  3x   3
The chain rule can be used more
than once.
(That’s what makes the “chain” in the
“chain rule”!)
6cos  3x  sin  3x 

𝒅
𝒇 𝒖
𝒅𝒙
𝒅𝒖
= 𝒇′(𝒖)
𝒅𝒙
Example 5
• Find
𝒅
[𝐬𝐢𝐧(𝟐𝒙)]
𝒅𝒙
• Solution. Taking u=2x in the generalized formula for 𝐬𝐢𝐧 𝒖
yields
𝒅
𝒅
𝒅𝒖
𝒅𝒙
𝒅𝒙
[𝒔𝒊𝒏(𝟐𝒙)]= [𝒔𝒊𝒏(𝒖)]=𝐜𝐨𝐬 𝒖
𝒅𝒙
=𝒄𝒐𝒔 𝟐𝒙
= 𝒄𝒐𝒔 𝟐𝒙 . 2 = 2𝒄𝒐𝒔 𝟐𝒙
𝒅
[2x]
𝒅𝒙
Example 7
Find
𝒅
𝒅𝒙
𝟑
𝟒
[x²−x+2]
Solution. Taking u = x²−x+2 in the generalized formula for
𝟑
𝟒
[𝒖] yields
𝒅
𝒅𝒙
𝟑
𝟒
[x²−x+2] =
𝒅
𝒅𝒙
𝟑
𝟒
[𝒖 ]
𝟑 −𝟏 𝒅𝒖
= 𝒖𝟒
𝟒
𝒅𝒙
−𝟏
𝟑
𝒅
= (x²−x+2) 𝟒 · [x²−x+2 ]
𝟒
𝒅𝒙
−𝟏
𝟑
= (x²−x+2) 𝟒
𝟒
(2x -1)
Select the best answer for the following question.
2x  h  5x  h  2x 2  5x
1. Find f x if f x  lim
.
h0
h
2




a)
f x  4x  5
b)
f x  2x  5x
c)
f x  2x 2  5x
d)
f x  2x  h  5x  h

2
2
Q
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2



2. Which of the following equations has derivative
dy
dx
 12x 3  24 x  6 ?
a) y  3x 4 12x 2  6x 11

b) y  36x 2  24
c) y  3x 4 12x 2  6
d) y 12x 4  24 x 2  6x 11
Q
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
Select the best answer for the following question.
Select the best answer for the following question.





4 16
x
, find f x.
a)
f x  36x  x
b)
f x  36x  x
c)
f x  36x  x
d)
f x  36 x 5
3
12
5
5
1
2
1
2
3
4
54
34
Q
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4
4
f
x

9x

3. For  



5 3  4
dy
4. For y    2 2t  6, find
.
dt
t t 
 5 6  3
dy
a)
  2  3 8t 
dt  t t  
 5 6 
dy 5 3  3
4
b)
   2 8t  2t  6 2  3 
 t t 
dt t t 

dy 5 3  3
6 
4
c)
   2 8t  2t  65  

dt t t 
t 
 5 6 
dy 5 3  3
4
   2 8t  2t  6 2  3 
d)
 t t 
dt t t 
Q
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
Select the best answer for the following question.



x3  6
5. For y 
, find y.
x3
2x 3  9x 2  6
a) y
2
x  3
3
2
4x

9x
6
b) y
2
x

3
 
c) y 2x
2x 3  9x 2  6
d) y
2
x

3
 
Q
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
Select the best answer for the following question.

6. For f x  3tan x  x cosx , find f x .
a) f x  3sec 2 x  sin x

b) f x  3sec x  xsin x  cosx
c) f x  3sec 2 x  x sin x  cos x


d) f x  3sec 2 x  x sin x  cos x
Q
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
Select the best answer for the following question.


7. Find an equation of the tangent line to the graph of
y  sin x 1 at x  2 .
a) y  2
b) y 
 x 1
c) y  0
Q
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8

Select the best answer for the following question.
d) y  x




8. For y  5x 14 x 
7
2
2
3
 , find
19

a) y 19x 5x 7 14 x 2  2
3



19
b) y 5x 14 x 
2
3


7
2
c) y 19 35x  28x
6

 35x 5  28
6
35x
  28x
18
d) y 19x 5x 14 x 
7

18
y.
2
2
3

18
5
35x
  28
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
Select the best answer for the following question.
Answers
5. a
1. c
6. d
2. a
7. c
3. b
8. d
4. d