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UNIT I-Module 6 Differentiation of trigonometric and Inverse trigonometric functions

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UNIT I - THE DERIVATIVES
Module 6: Differentiation of Trigonometric and Inverse Trigonometric Functions
Introduction
In this topic, we will discuss differentiation of new class of functions called the transcendental functions.
The trigonometric functions and their inverses, together with the logarithmic and exponential functions
are the simplest transcendental functions. This topic covers differentiation of trigonometric functions and
their inverses.
Learning Objectives
At the end of this topic, you should be able to demonstrate the following:
a. Determine the derivative of trigonometric functions; and
b. Find the derivative of inverse trigonometric functions.
Presentation of Contents
Differentiation of Trigonometric Functions
Let u be a differentiable function of x. That is u = g(x). The derivative of the six trigonometric functions
with respect to x are as follows:
1.
š‘‘
(sin š‘¢)
š‘‘š‘„
= cos š‘¢ . š‘‘š‘„
š‘‘š‘¢
2.
š‘‘
(cos š‘¢)
š‘‘š‘„
= −sin š‘¢ . š‘‘š‘„
3.
š‘‘
(tan š‘¢)
š‘‘š‘„
= š‘ š‘’š‘ 2 š‘¢ . š‘‘š‘„
š‘‘
š‘‘š‘¢
4. š‘‘š‘„ (cot š‘¢) = −š‘š‘ š‘ 2 š‘¢ . š‘‘š‘„
š‘‘š‘¢
5. š‘‘š‘„ (sec š‘¢) = sec š‘¢ tan š‘¢ . š‘‘š‘„
š‘‘
š‘‘š‘¢
š‘‘š‘¢
6. š‘‘š‘„ (csc š‘¢) = −csc š‘¢ cot š‘¢. š‘‘š‘„
š‘‘
š‘‘š‘¢
The following examples illustrate the use of the formulas in differentiating functions involving trigonometric
expressions.
EXAMPLE 1
a.
Find
š‘‘š‘¦
š‘‘š‘„
if y = csc(4š‘„ + 5).
Solution
š‘‘š‘¦
y = csc(4š‘„ + 5) takes the form of y = csc u with u = 4x + 5. Hence to find š‘‘š‘„, we use the formula
š‘‘š‘¦
š‘‘š‘„
=
š‘‘
(csc š‘¢)
š‘‘š‘„
š‘‘š‘¢
= −csc š‘¢ cot š‘¢. š‘‘š‘„
u = 4x + 5;
š‘‘š‘¦
š‘‘š‘„
š‘‘š‘¦
š‘‘š‘„
b.
š‘‘š‘¢
š‘‘š‘„
=4
= −csc(4š‘„ + 5) cot(4š‘„ + 5). (4)
= −4csc(4š‘„ + 5) cot(4š‘„ + 5)
Given f(x) = š‘ š‘–š‘›2 4š‘„ 2 , determine f ‘(x).
Solution
f (x) = š‘ š‘–š‘›2 4š‘„ 2 can be express as f (x) = (sin 4š‘„ 2 )2. This function takes the form of f (x) = š‘¢š‘› with
u = sin 4š‘„ 2 and n = 2. This suggests the use of the GPR, that is
š‘‘š‘¢
f ‘(x) = nš‘¢š‘›−1. š‘‘š‘„
u = sin 4š‘„ 2 and n = 2, gives
f ‘(x) = 2 (sin 4š‘„ 2 )2−1.
š‘‘
š‘‘š‘„
(sin 4š‘„ 2 );
we further differentiate š‘ š‘–š‘› 4š‘„ 2 using
š‘‘
(sin š‘¢)
š‘‘š‘„
š‘‘š‘¢
with u = 4š‘„ 2 , we have
= cos š‘¢ . š‘‘š‘„
š‘‘
(sin 4š‘„ 2 )
š‘‘š‘„
š‘‘
= cos 4š‘„ 2 . š‘‘š‘„ (4š‘„ 2 )
š‘‘
(sin 4š‘„ 2 )
š‘‘š‘„
= (cos 4š‘„ 2 ) (8x)
š‘‘
(sin 4š‘„ 2 )
š‘‘š‘„
= 8xcos 4š‘„ 2
f ‘(x) = 2( sin 4š‘„ 2 ) (8xcos 4š‘„ 2 )
f ‘(x) = 16x sin 4š‘„ 2 cos 4š‘„ 2
š‘‘š‘¦
1
Find š‘‘š‘„ if y = 2 x + 3tan 2š‘„
c.
Solution
1
š‘‘š‘¦
y = 2 x + 3tan 2š‘„ is a sum of two functions, so to determine š‘‘š‘„, we use the sum rule.
1
d. Find
y = 2 x + 3tan 2š‘„
differentiate both sides with respect to x
š‘‘š‘¦
š‘‘š‘„
= š‘‘š‘„ (2 x) + š‘‘š‘„ (3 tan 2š‘„)
using constant multiple rule
š‘‘š‘¦
š‘‘š‘„
=2
š‘‘š‘¦
š‘‘š‘„
= 2 (1) + 3 š‘‘š‘„ (tan 2š‘„)
š‘‘š‘¦
š‘‘š‘„
=
1
2
+ 3 [š‘ š‘’š‘2 2š‘„ .
š‘‘š‘¦
š‘‘š‘„
=
1
2
+ 3[(š‘ š‘’š‘2 2š‘„) (2) ]
š‘‘š‘¦
š‘‘š‘„
= 2 + 6š‘ š‘’š‘ 2 2š‘„
š‘‘š‘¦
š‘‘š‘„
=
š‘‘š‘¦
š‘‘š‘„
if sin(š‘„ + š‘¦) = x + y.
š‘‘
1
1 š‘‘
(x)
š‘‘š‘„
1
š‘‘
š‘‘
+ 3 š‘‘š‘„ (tan 2š‘„)
š‘‘
š‘‘
š‘‘š‘„
(2š‘„) ]
š‘‘
(š‘„)
š‘‘š‘„
=1
differentiate š‘”š‘Žš‘› 2š‘„ using
š‘‘
š‘‘š‘„
(š‘”š‘Žš‘› š‘¢) = š‘ š‘’š‘ 2 š‘¢ .
š‘‘š‘¢
š‘‘š‘„
š‘‘
but š‘‘š‘„ (2š‘„) = 2, so we have
simplifying further
1
1+12 š‘ š‘’š‘2 2š‘„
2
š‘‘š‘¦
The function is implicit, so we use implicit differentiation to obtain š‘‘š‘„.
sin(š‘„ + š‘¦) = x + y
differentiate both sides with respect to x gives
with u = 2x
š‘‘
[sin(š‘„
š‘‘š‘„
cos(š‘„ +
š‘‘
(x
š‘‘š‘„
+ š‘¦)] =
š‘‘
š‘¦) š‘‘š‘„ (x
+ y) =
š‘‘
cos(š‘„ + š‘¦) [š‘‘š‘„ (š‘„) +
cos(š‘„ + š‘¦) [ 1 +
+ y)
š‘‘š‘¦
]
š‘‘š‘„
to differentiate š‘ š‘–š‘›(š‘„ + š‘¦), use
š‘‘
š‘‘
(š‘„)
š‘‘š‘„
š‘‘
(y)]
š‘‘š‘„
š‘‘
(š‘¦)
š‘‘š‘„
+
š‘‘š‘„
š‘‘š‘¦
š‘‘š‘¦
š‘‘š‘„
š‘‘š‘„
;u=x+y
š‘‘š‘¦
š‘‘š‘¦
= 1 + š‘‘š‘„
Using DPMA on the left side of the equation
š‘‘š‘¦
š‘‘š‘¦
š‘‘š‘¢
= 1 + š‘‘š‘„
š‘‘š‘¦
cos(š‘„ + š‘¦) + cos(š‘„ + š‘¦) š‘‘š‘„ = 1 + š‘‘š‘„
cos(š‘„ + š‘¦) š‘‘š‘„ - š‘‘š‘„
(š‘ š‘–š‘› š‘¢) = š‘š‘œš‘  š‘¢ .
= 1 - cos(š‘„ + š‘¦)
[cos(š‘„ + š‘¦) − 1] = 1 - cos(š‘„ + š‘¦)
š‘‘š‘¦
š‘‘š‘„
=
1 − cos(š‘„+š‘¦)
cos(š‘„+š‘¦)−1
š‘‘š‘¦
š‘‘š‘„
=
−1[cos(š‘„+š‘¦)−1]
cos(š‘„+š‘¦)−1
š‘‘š‘¦
š‘‘š‘„
= -1
š‘‘š‘¦
collect terms with š‘‘š‘„ on one side of the equation
š‘‘š‘¦
factor out š‘‘š‘„ on one side of the equation
by MPE
simplify further by factoring out -1 in the numerator
Application
Activity 1
š‘‘š‘¦
Find š‘‘š‘„ and simplify your answer whenever possible.
1. y = sin š‘„ + cos š‘„ + tan š‘„
2. y = š‘ š‘’š‘ 4 4š‘„
3. y = š‘„ 2 cot 2x
1−š‘ š‘–š‘›š‘„
4. y = 1+cos š‘„
5. cos(š‘„š‘¦) = x – y
Differentiation of Inverse Trigonometric Functions
Let u be a differentiable function of x; u = g(x). The derivative of the six inverse trigonometric functions
with respect to x are as follows:
1.
2.
3.
4.
5.
š‘‘
š‘‘š‘„
š‘‘
š‘‘š‘„
š‘‘
š‘‘š‘„
š‘‘
š‘‘š‘„
š‘‘
š‘‘š‘„
(Arcsin š‘¢) =
(Arccos š‘¢) =
(Arctan š‘¢) =
(Arccot š‘¢) =
(Arcsec š‘¢) =
1
š‘‘š‘¢
√1− š‘¢2
š‘‘š‘„
−1
š‘‘š‘¢
√1− š‘¢2 š‘‘š‘„
1
š‘‘š‘¢
1+ š‘¢2 š‘‘š‘„
−1
š‘‘š‘¢
1+ š‘¢2
š‘‘š‘„
1
š‘‘š‘¢
š‘¢√š‘¢2 −1 š‘‘š‘„
š‘‘
6.
š‘‘š‘„
(Arccsc š‘¢) =
−1
š‘‘š‘¢
š‘¢√š‘¢2 −1
š‘‘š‘„
The notation š‘ š‘–š‘›−1 š‘„ used in trigonometry is considered inconvenient for this particular topic since it
might be read as “sin x with an exponent of -1” because -1 is not an exponent but it is the symbol for
1
inverse of sin. Therefore š‘ š‘–š‘›−1 š‘„ does not mean (š‘ š‘–š‘› š‘„)−1 or š‘ š‘–š‘› š‘„.
EXAMPLE 2
š‘‘š‘¦
a. Find š‘‘š‘„ if y = arccot š‘„ 2 .
y = arccot š‘„ 2 takes the form of y = arccot u with u = š‘„ 2 . Hence to find
š‘‘
(Arccot š‘¢)
š‘‘š‘„
−1
we use the formula
š‘‘š‘¢
= 1+ š‘¢2 š‘‘š‘„
y = arccot š‘„ 2 ; u = = š‘„ 2
−1
š‘‘š‘¦
,
š‘‘š‘„
š‘‘š‘¦
š‘‘š‘„
= 1+ (š‘„ 2 )2 š‘‘š‘„ (š‘„ 2 )
š‘‘š‘¦
š‘‘š‘„
= 1+š‘„4 (2x)
š‘‘š‘¦
š‘‘š‘„
= 1+š‘„4
by substitution
š‘‘
−1
−2š‘„
š‘‘š‘¦
b. Find š‘‘š‘„ if y = Arctan x + Arcsec √1 + š‘„ 2
š‘‘š‘¦
y = Arctan x + Arcsec √1 + š‘„ 2 is a sum of two functions, so to determine š‘‘š‘„, we use the sum rule.
y = Arctan x + Arcsec √1 + š‘„ 2
š‘‘š‘¦
š‘‘š‘„
š‘‘š‘¦
š‘‘š‘„
š‘‘
š‘‘
= š‘‘š‘„ (Arctan x) + š‘‘š‘„ (Arcsec √1 + š‘„2 )
1
š‘‘
1
= 1+ š‘„ 2 +
š‘‘š‘„
š‘‘š‘¦
š‘‘š‘¦
=
1
1+ š‘„ 2
+
1
= 1+ š‘„ 2 +
š‘‘š‘„
š‘‘š‘¦
š‘‘š‘„
š‘‘š‘¦
š‘‘š‘„
š‘‘š‘¦
š‘‘š‘„
š‘‘š‘¦
1
= 1+ š‘„ 2 +
√1+ š‘„2 √(√1+ š‘„2 ) −1
1
1
š‘„√1+ š‘„2
1
š‘„√1+ š‘„2
1
2
.
.
š‘„
1
(1+ š‘„2 )2
š‘„
1
(1+ š‘„2 )2
š‘„
š‘„√1+ š‘„2 √1+ š‘„2
1
1
š‘‘
. [ (1 + š‘„2 )−2 . 2š‘„ ]
2
1
1
1
. [2 (1 + š‘„2 )−2 . š‘‘š‘„ (1 + š‘„2 ) ]
1
š‘„√1+ š‘„
1
1
. š‘‘š‘„ (1 + š‘„2 )2
1
√1+ š‘„2 √š‘„2
. š‘‘š‘„ (√1 + š‘„2 )
š‘‘
√1+ š‘„2 √(1−š‘„2 ) −1
1
= 1+ š‘„ 2 +
š‘‘
2
1
1
= 1+ š‘„ 2 +
š‘‘š‘„
š‘‘š‘„
1
= 1+ š‘„ 2 . š‘‘š‘„ (š‘„) +
š‘‘š‘¦
š‘‘š‘¦
differentiate both sides of the equation with respect to x
= 1+ š‘„ 2 + 1+ š‘„2
= 1+ š‘„ 2 + 1+ š‘„2 =
š‘‘š‘„
2
1+ š‘„ 2
š‘‘š‘¦
c. Find š‘‘š‘„ if arcsin x + arcos y = y
š‘‘š‘¦
arcsin x + arcos y = y is implicit, so to obtain š‘‘š‘„, we use implicit differentiation.
arcsin x + arcos y = y
differentiate both sides of the equation with respect to x
š‘‘
š‘‘š‘¦
š‘‘š‘„
(arcsin x) +
1
š‘‘
š‘‘š‘„
š‘‘
√1− š‘„2
1
√1− š‘„2
1
1
š‘‘š‘¦
1
= š‘‘š‘„ +
1
š‘‘š‘¦
. š‘‘š‘„ (š‘¦) = š‘‘š‘„
š‘‘š‘¦
1
š‘‘š‘¦
√1− š‘¦2 š‘‘š‘„
= š‘‘š‘„ (1 +
1
√1− š‘¦ 2
)
√1− š‘¦2 +1
š‘‘š‘¦
= š‘‘š‘„ (
2
√1− š‘„
š‘‘š‘„
= š‘‘š‘„
š‘‘š‘¦
√1− š‘„2
š‘‘
√1− š‘¦ 2
√1− š‘¦ 2 š‘‘š‘„
š‘‘š‘¦
√1− š‘„2
−1
. š‘‘š‘„ (š‘„) +
-
(arcos y) =
√1− š‘¦ 2
)
1
š‘‘š‘¦
š‘‘š‘„
=
√1− š‘„2
√1− š‘¦2 +1
√1− š‘¦2
š‘‘š‘¦
1
= √1−
š‘‘š‘„
š‘‘š‘¦
=
š‘‘š‘„
š‘„2
.
√1− š‘¦2
√1− š‘¦ 2 +1
√1− š‘¦ 2
√1− š‘„ 2 (√1− š‘¦ 2 +1)
Application
Activity 2
š‘‘š‘¦
Find š‘‘š‘„ and simplify your answer whenever possible.
1. y = Arccot (tan 2x)
2. y = 3arcsin x – 2arctanx
tan š‘„
3. f(x) = arctan š‘„
4. f(x) = sin š‘„ arccosx
5. arccot x – tan y = 1 + y
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