Part of Mapua (MIT) Syllabus Content

1. DIFFERENTIATION OF
INVERSE TRIGONOMETRIC
FUNCTIONS

2. TRANSCENDENTAL FUNCTIONS
Kinds of transcendental functions:
1.logarithmic and exponential functions
2.trigonometric and inverse trigonometric
functions
3.hyperbolic and inverse hyperbolic functions
Note:
Each pair of functions above is an inverse to
each other.

3. The INVERSE TRIGONOMETRIC FUNCTIONS
.
x.issinewhoseangletheisymeanalsoThis
xsinyorxarcsinybydenoted
xoffunctionsineinversethecalledisyxysin
relationthebydeterminedxoffunctionaisyif
FunctionsricTrigonometInverseofPropertiesandsDefinition
callRe
1-
•
==
→=
•
-1xif0y
2
π
-or
1xifπ/2y0:wherexycscifx1cscy
-1xifyπ/2or
1xifπ/2y0:wherexysecifx1-secy
πy0:wherexycotifx1coty
π/2yπ/2-:wherexytanifx1tany
πy0:wherexcos yifx1cosy
π/2yπ/2-:wherexysinifx1siny
:sdefinitionfollowingthearethesegeneral,In
≤<≤
≥≤<===>−=
≤≤<
≥<≤===>=
<<===>−=
<<===>−=
≤≤===>−=
≤≤===>−=
π

4. DIFFERENTIATION FORMULA
Derivative of Inverse Trigonometric Function
( )
( )
functions.rictrigonomet
othertheforformulasthederivecanwemannersimilarIn
x-1
1
dx
xsind
xsinybut
x-1
1
dx
dy
x-1ysin-1ycos:identitythefrom
ycos
1
dx
dy
or
dy
dx
ycos
:ytorespectwithtingifferentiaD
2
y
2
-wherexysinfunction
rictrigonometinverseofdefinitiontheusewe,xsinyofderivativethefindingIn
2
1-
1-
2
22
-1
dx
du
u-1
1
usin
dx
d
Therefore
2
1-
=
=→==
==
==
≤≤=→
=
ππ

5. DIFFERENTIATION FORMULA
Derivative of Inverse Trigonometric Function
( )
( )
( )
( )
( )
( ) dx
du
1uu
1
ucsc
dx
d
6.
dx
du
1uu
1
usec
dx
d
5.
dx
du
u1
1
ucot
dx
d
4.
dx
du
u1
1
utan
dx
d
3.
dx
du
u1
1
ucos
dx
d
2.
dx
du
u1
1
usin
dx
d
1.
:functionsrictrigonometinverseforformulasationDifferenti
2
1
2
1
2
1
2
1
2
1
2
1
−
−=
−
=
+
−=
+
=
−
−=
−
=
−
−
−
−
−
−

6. A. Find the derivative of each of the following
functions and simplify the result:
( ) 31
xsinxf.1 −
=
( )
( )2
23
3x
x1
1
(x)f'
−
=
( ) 6
6
6
2
x1
x1
x1
3x
xf'
−
−
•
−
=
( ) ( )x3cosxf.2 1−
=
( ) 2
2
2
9x1
9x1
9x1
3
xf'
−
−
•
−
−
=
EXAMPLE:
( )
( )
( )3
3x1
1
xf'
2
−
−=
( ) 6
62
x1
x13x
xf'
−
−
= ( ) 2
2
9x1
9x13
xf'
−
−−
=
( ) 6
2
x1
3x
xf'
−
= ( ) 2
9x1
3
xf'
−
−
=

7. ( )21
x2secy.3 −
=
( )
( )4x
12x2x
1
y'
222
−
=
14xx
2
y'
4
−
=
xcos2y.4 1−
=
( )






⋅
−
−
⋅=
x2
1
x1
1
2'y
2
( )x
'y
−
−
=
⋅−
−
=
1x
1
xx1
1
14x
14x
14xx
2
y'
4
4
4
−
−
•
−
=
( )14xx
14x2
y' 4
4
−
−
=
( )
( )
( )x-1x
x-1x
x-1x
1
•
−
='y
( )
( )x-1x
x-1x−
='y

8. ( ) ( )x1
e2sin
2
1
xh.5 −
=
( )
( )2x
x
e21
e2
2
1
x'h
−
⋅=
x2
x2
x2
x
e41
e41
e41
e
−
−
•
−
=
( ) t5csct5sectg.6 11 −−
+=
( ) ( ) ( )5
125t5t
1)(
5
125t5t
1
tg'
22
−
−
+
−
=
( )
x
2
otcxg.7 1−
=
( ) 




 −






+
−
= 22
x
2
x
2
1
1
x'g
2
2
x
x
4
1
2
⋅





+
= ( )
4x
2
x'g 2
+
=→
x2
x2x
e41
e41e
−
−
=
( ) 0tg' =

9. ( ) ( )x3tanxxf.8 12 −
=
( )
( )
x2x3tan3
x31
1
xxf 1
2
2
•+





•
+
= −
( ) 





+
+
= −
x3tan2
x91
x3
xxf 1
2
)
x
5
(cscSecy.9 1−
=
1
x
5
csc
x
5
csc
x
5
x
5
cot
x
5
csc
'y
2
2
−









−−
=
x
5
cot
x
5
cot1
x
5
csc,but
22
=





=−





2
'
x
5
y =
( ) ( )( )






+
++
=
−
2
12
x91
x3tanx912x3
xxf

10. A. Find the derivative and simplify the result.
( ) x3tan3xg.1 1−
=
xcot
2
1
x2sinxy.2 11 −−
+=
( ) 3
1
x
4
sinxf.3 −
=
( )4
x2cscarcy.4 =
( ) x2Cosx5xG.5 12 −
=
( )xsincosy.6 1−
=
( )
x9
x3cot
xF.7
21−
=
( )x3tansiny.8 11 −−
=
( ) 211
xsecx6x3sinxh.9 −−
−=
21
5
x5cot
x7
y.10 −
=
EXERCISES:

11. ( )x2cos7y.4 1−
=
( )
2
t
arcsin4t4ttg.1 2
+−=
21
xcosy.2 −
−=
( ) z3secarczzf.3 4
=
( )x71tany.5 1
−= −
( ) ( )55
yarccosyyh.6 =








+
=
4x
x
arcsiny.7
2
( ) 





−
+
=
y1
y1
arctanyF.8
x4cosx4tany.9 11 −−
+=
( )
x4tan
4x
xH.10 1−
+
=
B. Find the derivative and simplify the result.