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2.6 derivatives of inverse hyperbolic functions

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Published by asyrafsafiyya, 2022-07-19 10:57:29

2.6 derivatives of inverse hyperbolic functions

2.6 derivatives of inverse hyperbolic functions

 −1 = 1  ℎ−1 = 1
1 − 2 1 + 2

 −1 =− 1  ℎ−1 = 1
1 − 2 2 − 1

 −1 1  ℎ−1 1
= 1 + 2 = 1 − 2

 −1 1  ℎ−1 1
= − 1 + 2 = 2 − 1

 −1 = 1  ℎ−1 =− 1
2 − 1 1 − 2

 −1 =− 1  ℎ−1 =− 1
2 − 1 1 + 2

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

.
.

Given = ℎ−1 4 . Differentiate wrt x.

From = ℎ−1 4

1 4= 4 #
= 1 + 4 2
1 + 16 2

ℎ−1 = 1
1 + 2 4 = 4

MOOC MAT438/ UiTM

.
.

Given = ℎ−1 . Differentiate wrt x.
2

From = ℎ−1
2

1 1 11 11 = 21
= 2 = 2 − 4 ∙ 2 = 2 − 4 ∙ 2 2 − 4 ∙ 2
2
2 42
−1

1 1 1
1 2 = 2 = 2 =#
2 − 1
2 − 4

ℎ−1 =

MOOC MAT438/ UiTM

.
.

Given = ℎ−1 2 . Differentiate wrt x.

From = ℎ−1 2 From =
1 → 2 2 = 2∙2 = 2 2
= 1 − 2 2
2 2 2 2
= 1 − 2 2 #

ℎ−1 1 = 2 = 2
= 1 − 2

MOOC MAT438/ UiTM

Given = ℎ−1 2 . Differentiate wrt x. Ans : = 2 2
1− 2 2

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

= ℎ−1
2

= = ℎ−1
2
Differentiate (using product rule),
Differentiate using composite function rule,
1 1
′ = 2 2 2
= = 2
′ = 1 1+
′ 2
=

′ = ′ + ′ ′ = 1 ∙ 2 1
2 2 2 ∙2

′ = + 1 ∙ 2 1 1
2 ∙2 = 2 2
′ =
2

MOOC MAT438/ UiTM

Solution :

From = ℎ−1

2

Differentiate implicitly wrt x
1

+ = 2 2

Multiplying each terms by 2 (to eliminate denominator 2)
1

2 + 2 = 2 2 2


2 + 2 = 2

Separating the term containing and not containing


2 = 2 − 2

Write as a subject,
2
− 2

= 2 #

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

3 = ℎ−13 + ℎ


= 3 = ℎ ℎ−−113 1 = ℎ
= 1 − 2
Differentiate (using ln rule), Differentiate using
Differentiate using product rule, exponential rule,

1 = = ℎ−13 ′ = ℎ ∙ ℎ
3
′ = ∙ 3 2 ′ = ′ = 1 ∙3
3
1 − 2

3 ′ = ′ + ′ Copy the
′ =
whole function

′ = ℎ−13 3 =
+ 1 − 9 2

Differentiate power

cosh = ℎ

MOOC MAT438/ UiTM

Solution :

From 3 = ℎ−13 + ℎ

Differentiate implicitly wrt x

3 ℎ−13 3 + ℎ ∙ ℎ
= + 1 − 9 2

Separating the term containing and not containing


3 3 − ℎ ℎ = ℎ−13
− 1 − 9 2

Write as a subject,

Multiplying each terms by 1 − 9 2 (to eliminate denominator and 1 − 9 2

3 − 3 − ℎ ℎ 1 − 9 2 3 1 − 9 2 3 − 1 − 9 2 ℎ ℎ = 1 − 9 2 ℎ−13
1 − 9 2 − 1 − 9 2

= ℎ−13 #

3 1 − 9 2 − 3 − 1 − 9 2 ℎ ℎ = 1 − 9 2 ℎ−13

3 1 − 9 2 − 3 − 1 − 9 2 ℎ ℎ #
=
1 − 9 2 ℎ−13

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM


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