Download Derivatives of Trigonometric Functions section 3.9

Derivatives of Trigonometric Functions
section 3.9
Consider the unit circle ; so we have OC = 1 in the picture below.
△
△
Because of similarity between the two triangles OCD and OAB we have
CD
AB
=
OC
OA
⇒
CD
sin θ
=
1
cos θ
⇒
CD = tan θ
D
B
θ
O
A
C
Because of the equality CD = tan θ , the axis that passes through the points C and D is called
the tangent axis.
Now assume for a moment that the radius of the circle is r. The sector OBC faces the polar
angle θ.
angle
area of associated sector
2π
πr2
θ
?
1
⇒
(θ)(πr2 )
1
= θr2
2π
2
?=
area of sector with polar angle θ = 12 θr2
Note: This formula is true only if θ is measured in radians.
Now let us assume for a moment that the above circle has radius 1. Then by putting r = 1 in
the above formula we have:
1
area of sector OBC = θ
2
Now note that
area of triangle OAB < area of sector OBC < area of triangle OCD
1
1
1
(OA)(AB) < θ < (OC)(CD)
2
2
2
1
1
1
(cos θ)(sin θ) < θ < (1)(tan θ)
2
2
2
multiply by 2 and divide by sin θ
cos θ <
θ
1
<
sin θ
cos θ
Now let θ → 0+ and use the Sandwich Theorem to get
lim
θ→0+
θ
=1
sin θ
Then
lim
θ→0+
sin θ
1
1
= lim θ =
+
θ
θ→0
limθ→0+
sin θ
2
θ
sin θ
=
1
=1
1
(∗)
Now we calculate the left-hand limit:
limθ→0−
sin θ
θ
= limx→0+
sin(−x)
−x
= limx→0+
− sin x
−x
= limx→0+
sin x
x
= 1
change of variable x = −θ
using the identity sin(−x) = − sin x
using the identity (*)
So ,
lim−
θ→0
sin θ
=1
θ
(∗∗)
From both (∗) and (∗∗) we have
sin θ
θ→0 θ
lim
=1
Example (section 3.9 exercise 33). Calculate lim
θ→0
tan θ
θ
Solution.
sin θ
tan θ
sin θ 1
= lim cos θ = lim
θ→0 θ
θ→0 θ
θ→0 θ
cos θ
lim
= lim
θ→0
sin θ
1
1
lim
= (1)( ) = 1
θ θ→0 cos θ
1
sin( x2 )
1
x→∞ sin( x )
Example (section 3.9 exercise 36). Calculate lim
3
Solution. Using the change of variable y =
1
x
we can write
sin( x2 )
sin(2y)
lim
= lim
= lim
1
x→∞ sin( )
y→0
y→0 sin(y)
x
sin(2y)
y
sin y
y
= lim
2 sin(2y)
2y
y→0
sin y
y
=
(2)(1)
=2
1
We now calculate the derivative of the sine function: For this we use
sin′ (x) =
=
=
sin(x+∆x)−sin x
∆x
∆x→0
lim
2
∆x→0 ∆x
lim
lim
∆x→0
[
sin
sin( ∆x
2 )
(
∆x
2
)
( ∆x )
2
(
cos x +
∆x
2
(
lim cos x +
∆x→0
)]
∆x
2
using
a+b
sin a − sin b = 2 sin( a−b
2 ) cos( 2 )
)
= (1) cos(x + 0) = cos x
So
d
dx
sin x = cos x
Now we find the derivative of the cosine function:
(π
)
d
d
cos
x
=
sin
−
x
dx
dx
2
= sin′
(π
= cos
(π
){
}′
− x π2 − x
2
2
= − cos
)
− x {−1}
(π
2
)
−x
= − sin x
So
d
dx
cos x = − sin x
4
Now the derivative of the tangent function :
tan′ x =
d sin x
dx cos x
=
{sin x}′ {cos x}−{sin x}{cos x}′
cos2 x
=
{cos x}{cos x}−{sin x}{− sin x}
cos2 x
=
cos2 x+sin2 x
cos2 x
d
dx
=
1
cos2 x
= sec2 x
tan x = sec2 x
Now the derivative of the cotangent function :
cot′ x =
d cos x
dx sin x
=
{cos x}′ {sin x}−{cos x}{sin x}′
sin2 x
=
{− sin x}{sin x}−{cos x}{cos x}
sin2 x
=
−(cos2 x+sin2 x)
sin2 x
d
dx
=
−1
sin2 x
cot x = − csc2 x
5
= − csc2 x
Now the derivative of the secant function :
sec′ x =
d
dx
(
1
cos x
)
=
{1}′ {cos x}−{1}{cos x}′
cos2 x
=
{0}{cos x}−{1}{− sin x}
cos2 x
=
sin x
cos2 x
d
dx
=
1 sin x
cos x cos x
= sec x tan x
sec x = sec x tan x
Now the derivative of the cosecant function :
csc′ x =
d 1
dx sin x
=
{1}′ {sin x}−{1}{sin x}′
sin2 x
=
{0}{sin x}−{1}{cos x}
sin2 x
=
− cos x
sin2 x
d
dx
= − sin1 x
cos x
sin x
= − csc x cot x
csc x = − csc x cot x
6
Putting all these together:
d
dx
sin x
d
dx
cos x =
− sin x
d
dx
tan x =
sec2 x
d
dx
cot x =
− csc2 x
d
dx
sec x =
sec x tan x
d
dx
csc x = − csc x cot x
=
cos x
Note: These formulae are true only if the unit of measurement is radian.
And , using the chain rule , we have the following formulae if interim variables are involved:
d
dx
sin u
(cos u) du
dx
d
dx
cos u =
(− sin u) du
dx
d
dx
tan u =
(sec2 u) du
dx
d
dx
cot u =
(− csc2 u) du
dx
d
dx
sec u =
(sec u tan u) du
dx
d
dx
csc u = (− csc u cot u) du
dx
=
7
Example. Find the equation of the tangent line to the curve y =
sec x
1+tan x
at the point where
x=0.
Solution.
y′ =
=
{sec x}′ {1 + tan x} − {sec x}{1 + tan x}′
(1 + tan x)2
{sec x tan x}{1 + tan x} − {sec x}{sec2 x}
(1 + tan x)2
⇒
x=0
slope =
(0)(1) − (1)(1)
= −1
(1)2
Also note that at x = 0 we have y = 1. Therefore:
tangent line
⇒
y − 1 = (−1)(x − 0)
dy
Example (from the textbook). Find dx for
Solution. Let u = 4x. Then
y = (tan u)
2
⇒
dy
dx
=
⇒
y = −x + 1
y = tan2 (4x)
{ }{ }
dy
du
du
dx
{
}{ }
du
= 2(tan u)(sec2 u) dx
{
}{ }
2
= 2(tan u)(sec u) 4
= 8 tan(4x) sec2(4x)
8
✓
power rule
Example (from the textbook - pages 203 and 204). Find
dy
dx
for x2 tan y + y sin x = 5
Solution. This is an implicit differentiation. If the “prime notation” is meant to denote differentiation with respect to x , then we have
{
} {
}
{x2 }′ {tan y} + {x2 }{tan y}′ + {y}′ {sin x} + {y}{sin x}′ = {5}′
} {
}
{
{2x}{tan y} + {x2 }{(sec2 y) y ′ } + {y ′ }{sin x} + {y}{cos x} = 0
factor out the term y′
⇒
⇒
y′ =
(2x tan y + y cos x) + (x2 sec2 y + sin x)y ′ = 0
−(2x tan y + y cos x)
(x2 sec2 y + sin x)
dy
Example (section 3.9 exercise 30 - modified). Find
dx for
✓
y =
1+tan3 (3x2 −4)
x2 sin(x3 )
.
Solution. Set u = 3x2 −4 and t = x3 . If the “prime notation” is meant to denote differentiation
9
with respect to x , then we have y =
dy
dx
=
1+(tan(u))3
x2 sin t
so then:
{1+(tan u)3 }′ {x2 sin t}−{1+(tan u)3 }{x2 sin t}′
(x2 sin t)2
{
3(tan u)2 (tan u)′
}{
} {
}{
=
x4 sin2 t
{
}{
3(tan u)2 (sec2 u)(u′ )
} {
}{
x2 sin t − 1+(tan u)3
=
{2x}{sin t}+{x2 }{(cos t)(t′ )}
}
x4 sin2 t
{
}{
} {
}{
x2 sin t − 1+(tan u)3
3(tan u)2 (sec2 u)(6x)
=
}
{2x}{sin t}+{x2 }{(cos t)(3x2 )}
x4 sin2 t
{
} {
18x3 tan2 u sec2 u sin t
=
{x2 }′ {sin t}+{x2 }{sin t}′
x2 sin t − 1+(tan u)3
}
−
}{
1+tan3 u
x4 sin2 t
where u = 3x2 − 4 and t = x3
10
2x sin t+3x4 cos t
}