2.3 derivatives of hyperbolic functions (eg1.2.3)

Derivatives of Trigonometric Functions Derivatives of Hyperbolic Functions


. = . =


. = − . =

. = . =



. = . = −


. = − . = −

. = − . = −


MOOC MAT438/ UiTM

= ℎ 2 .

Solution :

= ℎ 2

Differentiate wrt x,

= ℎ 2 2


2 = 2
ℎ = ℎ
#

= 2 ℎ 2

MOOC MAT438/ UiTM

= ℎ 3 .

Solution :

= ℎ 3

Differentiate wrt x,

− ℎ 3 ℎ( 3) 3 2
=

ℎ = − ℎ ℎ 3 = 3 2

= −3 2 ℎ 3 ℎ 3 #


MOOC MAT438/ UiTM

= ℎ .

Solution :

= ℎ

Differentiate wrt x,

− ℎ2 1
=
2

= 1
= − ℎ2 = 2
ℎ

ℎ2 = 1 −21
2
1
= − # =
2
2

MOOC MAT438/ UiTM

ℎ = 2 3 + −1 2



MOOC MAT438/ UiTM

ℎ = 2 3 + −1 2 .



Solution :

ℎ = 2 3 + −1 2 = −1 2
Let A = B + C
 ′ = 1 1
Differentiate, = 2 3
A’ = B’ + C’ 1 − 2 2

1
1 2 = 2 2
= ℎ −1 = 1 − 2 = 2 ∙ 1 −21

= 1′ = −21 = 1 2

′ = ℎ2 1 ′ = 2 3 3 1 − 4
′
ℎ = ℎ2 = ′ = 1
= 3 = 3
ℎ2 1
= − 4 2
= ′ = 6 3

MOOC MAT438/ UiTM

Solution :

Substitute A’, B’ and C’ into  :

A’ = B’ + C’ 

ℎ2 = 6 3 + 1
− 4 2

Multiplying each term by y

ℎ2 = ∙ 6 3 + ∙ 1
∙ − 4 2

ℎ2 6 3 + Standardize denominator for RHS :
= − 4 2
6 3 − 4 2
ℎ2 − 4 2 +
=
Write as a subject


6 3 + 6 3 − 4 2 +
= ℎ2 # = #
− 4 2 − 4 2 ℎ2


MOOC MAT438/ UiTM

ℎ32 + −1 = 2


MOOC MAT438/ UiTM

Example 5 : (implicit differentiation)

ℎ32 + −1 = 2


Solution :

ℎ32 + −1 = 2 −1 1
= 1 + 2

Let A +B =C = = −1

Differentiate, B’ = C’ = −1 ′ = 1 ′ = 1 1 ∙
A’ +  + 2

Using product rule, =

= ℎ32 = ℎ2 3 ′ = ′ + ′
= 2
1
Move the power to the front ′ = −1 1 + ∙ 1 + 2 ∙ ′ = 2 ∙ 2

′ = 3• ℎ2 2 ℎ2 2
′ −1 ′ = 2 2
+ 2
2 = 2 = + ∙
′ = 6 ℎ 2 2 ℎ = ℎ 2 ℎ #
1

MOOC MAT438/ UiTM

Solution :

Substitute A’, B’ and C’ into  :

A’ + B’ = C’ 

6 ℎ22 ℎ2 + −1 + = 2 2
1 + 2 ∙

Separating the term containing and not containing


6 ℎ22 ℎ2 + −1 = 2 2
− 1 + 2 ∙

6 ℎ22 ℎ2 + −1 2 2 −
= 1 + 2

Write as a subject, Factorize


6 ℎ22 ℎ2 + −1
= #
2 2 −
1 + 2

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM