Lesson 2 derivative of inverse trigonometric functions

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. is sine whoseangle the isy mean also This
xsiny or x arcsin yby denoted
x of function sineinverse the called isy x y sin
relation theby determined x of function a isy if
Functions ric Trigonomet Inverse of Properties and s Definition
callRe
1-
·
==
®=
·
-1x if 0 y
2
π
- or
1x if π/2 y0 :where x ycsc if x
1
cscy
-1x if yπ/2 or
1x if π/2 y0 :where x ysec if x
1-
secy
πy0 :where x ycot if x
1
coty
π/2yπ/2- :where x ytan if x
1
tany
πy0 :where x cos y if x
1
cosy
π/2yπ/2 - :where x ysin if x
1
siny
:sdefinition following the are these general, In
£<£
³£<===>
-
=
££<
³<£===>=
<<===>
-
=
<<===>
-
=
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-
=
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-
=
p

4. DIFFERENTIATION FORMULA
Derivative of Inverse Trigonometric Function
( )
( )
functions. ric trigonomet
other the for formulas the derive can wemanner similarIn
x-1
1
dx
xsind
xsiny but
x-1
1
dx
dy
x-1ysin-1y cos :identity the from
y cos
1
dx
dy
or
dy
dx
ycos
:y to respect withting ifferentiaD
2
y
2
- wherexy sin function
ric trigonomet inverse of definition the use we,xsiny of derivative the finding In
2
1-
1-
2
22
-1
dx
du
u-1
1
usin
dx
d
Therefore
2
1-
=
=®==
==
==
££=®
=
pp

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.
:functions ric trigonomet inverse for formulas ation Differenti
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
2
3
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'
2
22
-
=
14xx
2
y'
4
-
=
xcos2y .4
1-
=
( )
÷
ø
ö
ç
è
æ
×
-
-
×=
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
-
=
()
( )
2
x
x
e21
e2
2
1
x'h
-
×=
x2
x2
x2
x
e41
e41
e41
e
-
-
·
-
=
() t5csct5sec tg .6
11 --
+=
() () ()5
125t5t
1)(
5
125t5t
1
tg'
22
-
-
+
-
=
()
x
2
otc xg .7
1-
=
() ÷
ø
ö
ç
è
æ-
÷
ø
ö
ç
è
æ
+
-
=
22
x
2
x
2
1
1
x'g
2
2
x
x
4
1
2
×÷
ø
ö
ç
è
æ
+
= ()
4x
2
x'g
2
+
=®
x2
x2x
e41
e41e
-
-
=
()0tg'=

9. () ()x3tanxxf .8
12 -
=
()
()
x2x3tan3
x31
1
xxf
1
2
2
·+
÷
÷
ø
ö
ç
ç
è
æ
·
+
=
-
() ÷
ø
ö
ç
è
æ
+
+
=
-
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
-÷
ø
ö
ç
è
æ
ú
û
ù
ê
ë
é
--
=
x
5
cot
x
5
cot1
x
5
csc ,but
22
=÷
ø
ö
ç
è
æ
=-÷
ø
ö
ç
è
æ
2
'
x
5
y=
()
( )( )
÷
÷
ø
ö
ç
ç
è
æ
+
++
=
-
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 =
÷
÷
ø
ö
ç
ç
è
æ
+
=
4x
x
arcsiny .7
2
()
÷
÷
ø
ö
ç
ç
è
æ
-
+
=
y1
y1
arctanyF .8
x4cosx4tany .9
11 --
+=
()
x4tan
4x
xH .10
1-
+
=
B. Find the derivative and simplify the result.