Chapter 14- Trigonometric Functions
Chapter 15- Differentials and linear approximations
Lecture 24
MATH10070 - Introduction to Calculus
maths.ucd.ie/modules/math10070
Kevin Hutchinson
18th November 2010
Chapter 14- Trigonometric Functions
Chapter 15- Differentials and linear approximations
The derivatives of sin and cos
We have the following two fundamental formulae:
Z
d
sin x = cos x
dx
d
cos x = − sin x
dx
These formulae can be ‘guessed’ by examining the graphs of
sin and cos.
See the appendix to Chapter for a sketch of the proof.
Chapter 14- Trigonometric Functions
Chapter 15- Differentials and linear approximations
The four other trigonometric functions are derived from the two
basic ones by division.
So their derivatives can be found by using the quotient rule for
differentiation:
Chapter 14- Trigonometric Functions
Chapter 15- Differentials and linear approximations
The derivative of tan x.
d
sin x
dx cos x
cos x cos x − sin x(− sin x)
=
cos2 x
2
cos x + sin2 x
1
=
=
2
cos x
cos2 x
= sec2 x
d
(tan x) =
dx
Thus
Z
d
(tan x) = sec2 x
dx
Chapter 14- Trigonometric Functions
Chapter 15- Differentials and linear approximations
The derivative of sec x
Similarly,
d
1
dx cos x
cos x · 0 − 1 · (− sin x)
=
cos2 x
sin x
sin x
1
·
=
=
cos x cos x
cos2 x
= sec x tan x
d
(sec x) =
dx
Z
d
(sec x) = sec x tan x
dx
Chapter 14- Trigonometric Functions
Chapter 15- Differentials and linear approximations
Exercise: Show
d
(cot x) = − csc2 x
dx
d
(csc x) = − csc x cot x
dx
Chapter 14- Trigonometric Functions
Chapter 15- Differentials and linear approximations
More Examples
Example
Find
d
dx
sin(20x)
Solution: We must use the Chain Rule
d
sin(20x) = cos(20x) · 20 = 20 cos(20x).
dx
Chapter 14- Trigonometric Functions
Chapter 15- Differentials and linear approximations
Example
Find
d
dx
tan(x 3 ).
Solution: We must use the chain rule:
d
tan(x 3 ) = sec2 (x 3 ) · 3x 2
dx
= 3x 2 sec2 (x 3 )
Chapter 14- Trigonometric Functions
Chapter 15- Differentials and linear approximations
Example
Find
√
d
dx (csc(
x)).
Solution:
√
d
(csc( x)) =
dx
√
√
1
− csc( x) cot( x) · √
2 x
√
√
csc( x) cot( x)
√
= −
.
2 x