Calculus Notes Derivative of an Inverse Given f(a) = b → f

Commentaires

Transcription

Calculus Notes Derivative of an Inverse Given f(a) = b → f
Calculus Notes
Derivative of an Inverse
Given f(a) = b → f-1 (b) = a
f '(a) = m
The Derivative of the inverse at b is the reciprocal of the derivative of f(x) at a
1
1
1
(f-1 )'(b) = m
= f '(a)
=
f '(f-1 (b))
The Derivative of an inverse at a value is the reciprocal of the derivative of the Original at the pre-image.
Proof:
f(g(x)) = x
f '(g(x)) g'(x) = 1
1
g'(x) = (f-1 )'(x) = f '(g(x))
=
1
f '(f-1 (x))
Inverse Trig Functions:
y = arcsin x
y = arccos x
y = arctan x
y = arccot x = π2 - arctan x
y = arcsec x = arccos 1x
y = arccsc x = arcsin 1x
Derivatives of Inverse Trig Functions:
1.
y = arcsin x
sin y = x → cos y = ± 1 - x2
cos y • dy
=1
dx
dy
dx
1
= cos y =
d
(arcsin
dx
3.
1
1
1
=
2
sec y
1 + x2
1
du
d
(arctan
u)
=
2
dx
1 + u dx
y = arcsec x
sec y = x → tan y = ± x2 - 1
sec y tan y • dy
=1
dx
dy
dx
d
(arcsec
dx
u) =
1
|x|
1
|u|
x2 - 1
du
u2 - 1 dx
1
= - sin y = -
d
(arccos
dx
4.
u) = -
1
1 - x2
1
du
1 - u2 dx
y = arccot x
cot y = x → csc y = ± 1 + x2
-csc y • dy
=1
dx
1
1 + x2
1
du
d
(arccot
u)
=
2
dx
1 + u dx
dy
dx
=
1
= sec y tan y =
y = arccos x
cos y = x → sin y = ± 1 - x2
-sin y • dy
=1
dx
dy
dx
1 - x2
1
du
1 - u2 dx
y = arctan x
tan y = x → sec y = ± x2 + 1
sec2y • dy
=1
dx
dy
dx
5.
u) =
2.
6.
1
= - csc y = -
y = arccsc x
csc y = x → cot y = ± x2 - 1
-csc y cot y • dy
=1
dx
dy
dx
1
= - csc y cot y = -
d
(arccsc
dx
u) = -
1
|u|
1
|x|
x2 - 1
du
u2 - 1 dx
Integrals Resulting in Inverse Trig Functions
d
arcsin
dx
u=
1
du
1 - u2 dx
⌠ du
= arcsin u + C
⌡ 1 - u2
⌠
⌡ a2 - u2
du
du
a
=⌠

⌡
1-
u 2
a
()
()+C
= arcsin ua
--------------------------------------------------------d
arctan u = 1 2 du
dx
1 + u dx
∫1 + u
du
2
∫
= arctan u + C
⌠ a
1 + u 2
⌡ (a )
du
du
a2 + u2
a
=⌠
a1 + u 2
⌡  (a ) 
du
=
1
a
=
1
a
()+C
arctan ua
--------------------------------------------------------d
arcsec
dx
u=
1
|u|
du
u2 - 1 dx
du
⌠
= arcsec u + C
⌡|u| u2 - 1
⌠
⌡|u| u2 - a2
du
=⌠
|u|
⌡
du
a
u
2-1
a
()
=
⌠
u
⌡|a |
1
a
du
a
u
2-1
a
()
=
1
a
|u|
arcsec a
()
Summary:
Inverse Trigonometric Functions: Differentiation and Integration
∫
a2 - u2
• du
dx
∫
a2 - u2
[arcsin u] =
d
dx
[arccos u] =
3.
d
dx
[arctan u] = 1 +1 u2
• du
dx
∫a
4.
d
dx
[arccot u] = 1 +-1u2
• du
dx
∫ a -du+ u
5.
d
dx
[arcsec u] =
d
dx
[arccsc u] =
2.
6.
1
• du
dx
d
dx
1.
1 - u2
-1
1 - u2
1
|u|
u2 - 1
-1
|u|
u2 - 1
du
= arcsin ua + C
-du
du
2 + u2
2
• du
dx
∫u
• du
dx
∫u
2
= arccos ua + C
= 1a arctan ua + C
= 1a arccot ua + C
du
u2 - a2
-du
u2 - a2
|u|
= 1a arcsec a + C
|u|
= 1a arccsc a + C

Documents pareils

Diff and Integ Rules Final

Diff and Integ Rules Final  csc u cot u du   csc u  C  tan u du  ln sec u  C   ln cos u  C  cot u du   ln csc u  C  ln sin u  C  sec u du  ln sec u  tan u  C   ln sec u  tan u  C  csc u du   ln csc...

Plus en détail

DERIVATIVES AND INTEGRALS

DERIVATIVES AND INTEGRALS cos u   sin uu dx d sec u  sec u tan uu dx d u arccos u  dx 1  u2 d u arcsec u  dx u u2  1 d cosh u  sinh uu dx d sech u   sech u tanh uu dx d u cosh1 u ...

Plus en détail

Développements en séries entières usuels

Développements en séries entières usuels ( −1) ( 2n )! x 2 n = 1 − x 2 + 3x 4 − 5 x 6 + ...

Plus en détail

Algebra Geometry - Stewart Calculus

Algebra Geometry - Stewart Calculus y csc u cot u du  csc u  C 12. y tan u du  ln  sec u   C 13. y cot u du  ln  sin u   C 14. y sec u du  ln  sec u  tan u   C 15. y csc u du  ln  csc u  cot u   C

Plus en détail

Preparation for Exam 1

Preparation for Exam 1 (E) (integrations with trigonometric functions) to be very familiar with the differentiation formulas related to all six trigonometric functions and with the double angle/half angle formulas, and t...

Plus en détail