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Section 3.3 Derivatives of Trigonometric Functions
2015 Kiryl Tsishchanka
Derivatives of Trigonometric Functions
THE DERIVATIVE OF THE SINE AND COSINE FUNCTIONS: We have
d
(sin x) = cos x
dx
or
(sin x)′ = cos x
d
(cos x) = − sin x
dx
or
(cos x)′ = − sin x
and
Proof: We have
sin(x + h) − sin x
h→0
h
(sin x)′ = lim
[We use sin(α + β) = sin α cos β + cos α sin β]
sin x cos h − sin x cos x sin h
sin x cos h + cos x sin h − sin x
= lim
+
= lim
h→0
h→0
h
h
h
sin x(cos h − 1)
cos h − 1
sin h
sin h
= lim
= sin x lim
+ cos x ·
+ cos x lim
h→0
h→0
h→0 h
h
h
h
= sin x · 0 + cos x · 1 = cos x
In the same way we prove that (cos x)′ = − sin x.
EXAMPLE: If f (x) = 3 sin x − 4 cos x, then
f ′ (x) = (3 sin x − 4 cos x)′ = 3(sin x)′ − 4(cos x)′ = 3 cos x − 4(− sin x) = 3 cos x + 4 sin x
EXAMPLE: If f (x) = sin x, then
f ′ (x) = cos x
f ′′ (x) = (cos x)′ = − sin x
f ′′′ (x) = (− sin x)′ = − cos x
f ′′′′ (x) = (− cos x)′ = −(− sin x) = sin x
Therefore
f (x) = f (4) (x) = f (8) (x) = f (12) (x) = f (16) (x) = . . . = sin x
f ′ (x) = f (5) (x) = f (9) (x) = f (13) (x) = f (17) (x) = . . . = cos x
f ′′ (x) = f (6) (x) = f (10) (x) = f (14) (x) = f (18) (x) = . . . = − sin x
f ′′′ (x) = f (7) (x) = f (11) (x) = f (15) (x) = f (19) (x) = . . . = − cos x
For instance, it immediately follows from here that
f (2010) (x) = f (2008+2) (x) = f (4·502+2) = f ′′ (x) = − sin x
1
Section 3.3 Derivatives of Trigonometric Functions
2015 Kiryl Tsishchanka
DIFFERENTIATION RULES:
c′ = 0
(xn )′ = nxn−1
[cf (x)]′ = cf ′ (x)
(sin x)′ = cos x
(csc x)′ = − csc x cot x
[f (x) ± g(x)]′ = f ′ (x) ± g ′ (x)
(cos x)′ = − sin x (sec x)′ = sec x tan x
(tan x)′ = sec2 x
(cot x)′ = − csc2 x
[f (x)g(x)]′ = f ′ (x)g(x) + f (x)g ′ (x)
′
f (x)
f ′ (x)g(x) − f (x)g ′ (x)
=
g(x)
[g(x)]2
sec x
(sec x)′
???
′
EXAMPLE: If f (x) =
, then f (x) =
WRONG!!!
1 + tan x
(1 + tan x)′
′
sec x
(sec x)′ (1 + tan x) − sec x(1 + tan x)′
′
f (x) =
=
1 + tan x
(1 + tan x)2
=
sec x tan x(1 + tan x) − sec x sec2 x
sec x tan x(1 + tan x) − sec x(1′ + (tan x)′ )
=
(1 + tan x)2
(1 + tan x)2
sec x(tan x + tan2 x − sec2 x)
sec x(tan x − 1)
=
= [tan2 x − sec2 x = −1] =
2
(1 + tan x)
(1 + tan x)2
EXAMPLES: Differentiate
1. f (x) = 3x − 5
2. f (x) =
√
x
1
3. f (x) = √
x
√
4. f (x) = −3x−8 + 2 x
5. f (x) = (x3 + 7x2 − 8)(2x−3 + x−4 )
6. f (x) =
3x + 2
x+1
(x−5 + 1)
7. f (x) = x3 sin x − 5 cos x
8. f (x) =
sin x + sec x
1 + x tan x
9. f (x) =
sin x
1 + cos x
2
Section 3.3 Derivatives of Trigonometric Functions
2015 Kiryl Tsishchanka
Solutions:
1. (3x − 5)′ = (3x)′ − 5′ = 3x′ − 5′ = 3 · 1 − 0 = 3
′ 1
√
1
2. ( x)′ = x1/2 = x1/2−1 = x−1/2
2
2
′
′ ′
1
1
1
1
3. √
= x−1/2 = − x−1/2−1 = − x−3/2
=
1/2
x
2
2
x
√
1
4. (−3x−8 + 2 x)′ = −3(x−8 )′ + 2(x1/2 )′ = −3(−8)x−8−1 + 2 x−1/2 = 24x−9 + x−1/2
2
51 . [(x3 + 7x2 − 8)(2x−3 + x−4 )]′ = (x3 + 7x2 − 8)′ (2x−3 + x−4 ) + (x3 + 7x2 − 8)(2x−3 + x−4 )′
= (3x2 + 14x)(2x−3 + x−4 ) + (x3 + 7x2 − 8)(−6x−4 − 4x−5 )
52 . [(x3 + 7x2 − 8)(2x−3 + x−4 )]′ = (2 + 15x−1 + 7x−2 − 16x−3 − 8x−4 )′
= 2′ + 15(x−1 )′ + 7(x−2 )′ − 16(x−3 )′ − 8(x−4 )′
= 0 + 15 · (−1)x−2 + 7 · (−2)x−3 − 16 · (−3)x−4 − 8 · (−4)x−5
= −15x−2 − 14x−3 + 48x−4 + 32x−5
′ ′
3x + 2
3x + 2
3x + 2
−5
−5
(x + 1) =
(x−5 + 1)′
(x + 1) +
x+1
x+1
x+1
3x + 2
(3x + 2)′ (x + 1) − (3x + 2)(x + 1)′ −5
(x + 1) +
(x−5 + 1)′
=
(x + 1)2
x+1
3(x + 1) − (3x + 2) −5
x−5 + 1
3x + 2
3x + 2
−6
=
(−5x ) =
(−5x−6 )
(x + 1) +
+
(x + 1)2
x+1
(x + 1)2
x+1
6.
7. (x3 sin x − 5 cos x)′ = (x3 sin x)′ − 5(cos x)′ = (x3 )′ sin x + x3 (sin x)′ − 5(cos x)′
= 3x2 sin x + x3 cos x + 5 sin x
8.
9.
sin x + sec x
1 + x tan x
sin x
1 + cos x
′
′
=
(sin x + sec x)′ (1 + x tan x) − (sin x + sec x)(1 + x tan x)′
(1 + x tan x)2
=
[(sin x)′ + (sec x)′ ](1 + x tan x) − (sin x + sec x)(x′ tan x + x(tan x)′ )
(1 + x tan x)2
=
(cos x + sec x tan x)(1 + x tan x) − (sin x + sec x)(tan x + x sec2 x)
(1 + x tan x)2
=
cos x(1 + cos x) − sin x(− sin x)
(sin x)′ (1 + cos x) − sin x(1 + cos x)′
=
(1 + cos x)2
(1 + cos x)2
=
cos x + 1
1
cos x + cos2 x + sin2 x
=
=
2
2
(1 + cos x)
(1 + cos x)
1 + cos x
3
Section 3.3 Derivatives of Trigonometric Functions
2015 Kiryl Tsishchanka
Appendix
cos h − 1
0 A
(cos h − 1)(cos h + 1) A
cos2 h − 1
lim
=
= lim
= lim
h→0
h→0
h→0 h(cos h + 1)
h
0
h(cos h + 1)
A
−(1 − cos2 h)
h→0 h(cos h + 1)
= lim
T
− sin2 h
h→0 h(cos h + 1)
= lim
A
= lim
h→0
C
sin h − sin h
·
h
cos h + 1
sin h
− sin h
· lim
h→0 h
h→0 cos h + 1
= lim
− sin h
h→0 cos h + 1
C
= 1 · lim
DSP
=
=
− sin 0
1 + cos 0
0
1+1
=0
4