y - Penn Math
3/5/2012
Math 103 – Rimmer 3.9Inverse Trig. Functions
f ( x ) = arcsin x
or f ( x ) = sin −1 x
sin −1 x ≠
1 sin x
arcsin ( sin x ) = x sin ( arcsin x ) = x
Math 103 – Rimmer 3.9Inverse Trig. Functions
−π π Definition : y = arcsin x is the number in , for which sin y = x 2 2 3 π Find arcsin = 3 2 3 when plugged into the sine function. 2 π π 3 Find the angle b / w − and on the unit circle that goes with a point with a y − value of . 2 2 2
Find the angle that gives
−1 −π Find arcsin = 6 2 π π −1 Find the angle b / w − and on the unit circle that goes with a point with a y − value of . 2 2 2
Find arcsin ( −1) =
−π 2
π π Find the angle b / w − and on the unit circle that goes with a point with a y − value of − 1. 2 2
1
3/5/2012
What is the derivative of y = arcsin x? from before: f ( x ) = sin x 1 ′ ( f −1 ) ( x ) = f ′ f −1 ( x ) f ′ ( x ) = cos x ( )
1 ( f )′ ( x ) = cos ( arcsin x) 1
−1
( f )′ ( x ) = ′ x = ( )
(f )
( )( )
this is better written as
1 − sin ( arcsin x ) 2
1
−1
−1
f −1 ( x ) = arcsin x f −1 ′ x = ?
since cos x = 1 − sin 2 x , we have
−1
( f )′ ( x ) =
Math 103 – Rimmer 3.9Inverse Trig. Functions
remember sin ( arcsin x ) = x
1 − sin ( arcsin x )
2
y = arcsin x
1
y′ =
1 − x2
1 1 − x2
Math 103 – Rimmer 3.9Inverse Trig. Functions
Let y = arcsin ( 3x ) . 3 Find y ′ . 6
y′ =
1 1 − ( 3x )
2
⋅ ( 3)
3⋅
3 = 6
3 2
2
3 3 3 = y ′ = 1 3 6 1− 4 4 3 3 y ′ = 6 1 2
3 3 = 4 2
1 1 = 2 4
3 y ′ = 6 6
2
3/5/2012
Math 103 – Rimmer 3.9Inverse Trig. Functions
f ( x ) = arccos x or f ( x ) = cos −1 x
cos −1 x ≠
1 cos x
arccos ( cos x ) = x cos ( arccos x ) = x
Math 103 – Rimmer 3.9Inverse Trig. Functions
Definition : y = arccos x is the number in [ 0, π ] for which cos y = x
−1 2π Find arccos = 3 2 Find the angle ( b / w 0 and π ) on the unit circle that goes with a point with a x − value of
−1 . 2
3 π Find arccos = 6 2 Find the angle ( b / w 0 and π ) on the unit circle that goes with a point with a x − value of
3 . 2
Find arccos ( −1) = π Find the angle ( b / w 0 and π ) on the unit circle that goes with a point with a x − value of − 1.
3
3/5/2012
What is the derivative of y = arccos x? from before: f ( x ) = cos x 1 ′ f −1 ( x ) = f ′ ( x ) = − sin x f ′ f −1 ( x )
( )
(
)
1 ( f )′ ( x ) = − sin ( arccos x) 1
−1
f −1 ( x ) = arccos x f −1 ′ x = ?
( )( )
since sin x = 1 − cos 2 x , we have
−1
( f )′ ( x ) =
Math 103 – Rimmer 3.9Inverse Trig. Functions
− 1 − cos ( arccos x )
this is better written as
2
( f )′ ( x ) =
−1
−1
remember cos ( arccos x ) = x
1 − cos ( arccos x )
2
y = arccos x
( f )′ ( x ) =
−1
−1
y′ =
1 − x2
Fall 2010
y′ =
−1 ⋅ x −3/ 2 2 1 1− x
⇒ y′ ( 4 ) =
−1
2
−1 ⋅ 1 2 ⋅ 43/ 2 1− 4 −1
( )
−1 1 − x2
g=
1 = x −1/ 2 x
g′ =
−1 −3/2 x 2
2
1 1 since = x x
y′ =
Math 103 – Rimmer 3.9Inverse Trig. Functions
−1 ⋅ 3/2 1 2x 1− x −1
3/ 2 −1 −1 = ⋅ since 43/ 2 = 2 2 = 8 3 2 ⋅8 4 =
1 −1 −1 ⋅ = 3 16 8 3 2
y′ ( 4 ) =
1 8 3
4
3/5/2012
f ( x ) = arctan x or f ( x ) = tan x −1
lim arctan x =
x→− ∞
Math 103 – Rimmer 3.9Inverse Trig. Functions
−π 2
lim arctan x = x →∞
1 tan −1 x ≠ tan x
π 2
arctan ( tan x ) = x tan ( arctan x ) = x
Math 103 – Rimmer 3.9Inverse Trig. Functions
−π π Definition : y = arctan x is the number in , for which tan y = x 2 2
Find arctan ( −1) =
−π 4
tan ( π2 ) = undef tan ( π3 ) = 3
tan ( π4 ) = 1
Find arctan
( 3 ) = π3
1 3
tan ( π6 ) =
tan ( 0 ) = 0
−1 −π Find arctan = 6 3
tan ( −6π ) =
−1 3
tan ( −4π ) = −1
tan ( −3π ) = − 3 tan ( −2π ) = undef
5
3/5/2012
What is the derivative of y = arctan x? from before: f ( x ) = tan x 1 ′ f −1 ( x ) = f ′ ( x ) = sec 2 x f ′ f −1 ( x )
( )
(
)
1 ( f )′ ( x ) = sec ( arctan x)
Math 103 – Rimmer 3.9Inverse Trig. Functions
f −1 ( x ) = arctan x f −1 ′ x = ?
( )( )
since sec2 x = 1 + tan 2 x, we have
−1
2
( f )′ ( x ) = 1 + tan (1arctan x ) −1
this is better written as
2
( f )′ ( x ) =
1
−1
1 + tan ( arctan x )
remember tan ( arctan x ) = x 2
y = arctan x ′ x = ( )
(f ) −1
1 1 + x2
Why is the word arc used?
y′ =
1 1 + x2
Math 103 – Rimmer 3.9Inverse Trig. Functions
arclentgh s = rθ
On the unit circle r = 1,so s = θ
6
Math 103 – Rimmer 3.9Inverse Trig. Functions
f ( x ) = arcsin x
or f ( x ) = sin −1 x
sin −1 x ≠
1 sin x
arcsin ( sin x ) = x sin ( arcsin x ) = x
Math 103 – Rimmer 3.9Inverse Trig. Functions
−π π Definition : y = arcsin x is the number in , for which sin y = x 2 2 3 π Find arcsin = 3 2 3 when plugged into the sine function. 2 π π 3 Find the angle b / w − and on the unit circle that goes with a point with a y − value of . 2 2 2
Find the angle that gives
−1 −π Find arcsin = 6 2 π π −1 Find the angle b / w − and on the unit circle that goes with a point with a y − value of . 2 2 2
Find arcsin ( −1) =
−π 2
π π Find the angle b / w − and on the unit circle that goes with a point with a y − value of − 1. 2 2
1
3/5/2012
What is the derivative of y = arcsin x? from before: f ( x ) = sin x 1 ′ ( f −1 ) ( x ) = f ′ f −1 ( x ) f ′ ( x ) = cos x ( )
1 ( f )′ ( x ) = cos ( arcsin x) 1
−1
( f )′ ( x ) = ′ x = ( )
(f )
( )( )
this is better written as
1 − sin ( arcsin x ) 2
1
−1
−1
f −1 ( x ) = arcsin x f −1 ′ x = ?
since cos x = 1 − sin 2 x , we have
−1
( f )′ ( x ) =
Math 103 – Rimmer 3.9Inverse Trig. Functions
remember sin ( arcsin x ) = x
1 − sin ( arcsin x )
2
y = arcsin x
1
y′ =
1 − x2
1 1 − x2
Math 103 – Rimmer 3.9Inverse Trig. Functions
Let y = arcsin ( 3x ) . 3 Find y ′ . 6
y′ =
1 1 − ( 3x )
2
⋅ ( 3)
3⋅
3 = 6
3 2
2
3 3 3 = y ′ = 1 3 6 1− 4 4 3 3 y ′ = 6 1 2
3 3 = 4 2
1 1 = 2 4
3 y ′ = 6 6
2
3/5/2012
Math 103 – Rimmer 3.9Inverse Trig. Functions
f ( x ) = arccos x or f ( x ) = cos −1 x
cos −1 x ≠
1 cos x
arccos ( cos x ) = x cos ( arccos x ) = x
Math 103 – Rimmer 3.9Inverse Trig. Functions
Definition : y = arccos x is the number in [ 0, π ] for which cos y = x
−1 2π Find arccos = 3 2 Find the angle ( b / w 0 and π ) on the unit circle that goes with a point with a x − value of
−1 . 2
3 π Find arccos = 6 2 Find the angle ( b / w 0 and π ) on the unit circle that goes with a point with a x − value of
3 . 2
Find arccos ( −1) = π Find the angle ( b / w 0 and π ) on the unit circle that goes with a point with a x − value of − 1.
3
3/5/2012
What is the derivative of y = arccos x? from before: f ( x ) = cos x 1 ′ f −1 ( x ) = f ′ ( x ) = − sin x f ′ f −1 ( x )
( )
(
)
1 ( f )′ ( x ) = − sin ( arccos x) 1
−1
f −1 ( x ) = arccos x f −1 ′ x = ?
( )( )
since sin x = 1 − cos 2 x , we have
−1
( f )′ ( x ) =
Math 103 – Rimmer 3.9Inverse Trig. Functions
− 1 − cos ( arccos x )
this is better written as
2
( f )′ ( x ) =
−1
−1
remember cos ( arccos x ) = x
1 − cos ( arccos x )
2
y = arccos x
( f )′ ( x ) =
−1
−1
y′ =
1 − x2
Fall 2010
y′ =
−1 ⋅ x −3/ 2 2 1 1− x
⇒ y′ ( 4 ) =
−1
2
−1 ⋅ 1 2 ⋅ 43/ 2 1− 4 −1
( )
−1 1 − x2
g=
1 = x −1/ 2 x
g′ =
−1 −3/2 x 2
2
1 1 since = x x
y′ =
Math 103 – Rimmer 3.9Inverse Trig. Functions
−1 ⋅ 3/2 1 2x 1− x −1
3/ 2 −1 −1 = ⋅ since 43/ 2 = 2 2 = 8 3 2 ⋅8 4 =
1 −1 −1 ⋅ = 3 16 8 3 2
y′ ( 4 ) =
1 8 3
4
3/5/2012
f ( x ) = arctan x or f ( x ) = tan x −1
lim arctan x =
x→− ∞
Math 103 – Rimmer 3.9Inverse Trig. Functions
−π 2
lim arctan x = x →∞
1 tan −1 x ≠ tan x
π 2
arctan ( tan x ) = x tan ( arctan x ) = x
Math 103 – Rimmer 3.9Inverse Trig. Functions
−π π Definition : y = arctan x is the number in , for which tan y = x 2 2
Find arctan ( −1) =
−π 4
tan ( π2 ) = undef tan ( π3 ) = 3
tan ( π4 ) = 1
Find arctan
( 3 ) = π3
1 3
tan ( π6 ) =
tan ( 0 ) = 0
−1 −π Find arctan = 6 3
tan ( −6π ) =
−1 3
tan ( −4π ) = −1
tan ( −3π ) = − 3 tan ( −2π ) = undef
5
3/5/2012
What is the derivative of y = arctan x? from before: f ( x ) = tan x 1 ′ f −1 ( x ) = f ′ ( x ) = sec 2 x f ′ f −1 ( x )
( )
(
)
1 ( f )′ ( x ) = sec ( arctan x)
Math 103 – Rimmer 3.9Inverse Trig. Functions
f −1 ( x ) = arctan x f −1 ′ x = ?
( )( )
since sec2 x = 1 + tan 2 x, we have
−1
2
( f )′ ( x ) = 1 + tan (1arctan x ) −1
this is better written as
2
( f )′ ( x ) =
1
−1
1 + tan ( arctan x )
remember tan ( arctan x ) = x 2
y = arctan x ′ x = ( )
(f ) −1
1 1 + x2
Why is the word arc used?
y′ =
1 1 + x2
Math 103 – Rimmer 3.9Inverse Trig. Functions
arclentgh s = rθ
On the unit circle r = 1,so s = θ
6
