1. Hyperbolic functions

2. FM Hyperbolic functions: Calculus
KUS objectives
BAT differentiate and integrate with hyperbolic functions
Starter: differentiate
=−3 sin (2𝜋−3𝑥)
cos (2𝜋−3𝑥)
=8 cos 8𝑥
sin 8𝑥
= 𝑠𝑒𝑐 2 𝑥 2
tan 𝑥 2

3. d dx sinh 𝑥 = cosh 𝑥 𝑑 𝑑𝑥 cosh 𝑥 = sinh 𝑥 d dx tanh 𝑥 = 𝑠𝑒𝑐ℎ 2 𝑥
Notes
d dx sinh 𝑥 = cosh 𝑥
𝑑 𝑑𝑥 cosh 𝑥 = sinh 𝑥
d dx tanh 𝑥 = 𝑠𝑒𝑐ℎ 2 𝑥
Positive sign
d dx arsinh 𝑥 = 1 𝑥 2 +1
𝑑 𝑑𝑥 arcosh 𝑥 = 1 𝑥 2 −1 ,
d dx artanh 𝑥 = 1 1− 𝑥 2
𝑥>1
𝑥 <1

4. WB D1 Prove each result: 𝑎) d dx sinh 𝑥 = cosh 𝑥
𝑏) 𝑑 𝑑𝑥 cosh 𝑥 = sinh 𝑥 𝑐) d dx tanh 𝑥 = 𝑠𝑒𝑐ℎ 2
sinh 𝑥 ≡ 𝑒 𝑥 − 𝑒 −𝑥 2
cosh 𝑥 ≡ 𝑒 𝑥 + 𝑒 −𝑥 2
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
𝑎) sinh 𝑥 ≡ 𝑒 𝑥 − 𝑒 −𝑥 2 = 𝑒 𝑥 2 − 𝑒 −𝑥 2
𝑑 𝑑𝑥 sinh 𝑥 = 1 2 𝑒 𝑥 −(−1) 𝑒 −𝑥 = 1 2 𝑒 𝑥 𝑒 −𝑥 = cosh 𝑥
𝑏) cosh 𝑥 ≡ 𝑒 𝑥 + 𝑒 −𝑥 2 = 𝑒 𝑥 𝑒 −𝑥 2
𝑑 𝑑𝑥 sinh 𝑥 = 1 2 𝑒 𝑥 +(−1) 𝑒 −𝑥 = 1 2 𝑒 𝑥 − 1 2 𝑒 −𝑥 = sinh 𝑥
𝑐) tanh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
Use quotient rule
𝑑 𝑑𝑥 tanh 𝑥 = 𝑒 2𝑥 𝑒 2𝑥 − 𝑒 2𝑥 −1 2𝑒 2𝑥 𝑒 2𝑥 +1 2
𝑑 𝑑𝑥 tanh 𝑥 = 2𝑒 4𝑥 + 2𝑒 2𝑥 − 2𝑒 4𝑥 + 2𝑒 2𝑥 𝑒 2𝑥 = 4𝑒 2𝑥 𝑒 2𝑥 +1 2
= 2𝑒 𝑥 𝑒 2𝑥 = 2𝑒 𝑥 𝑒 2𝑥
divide through by 𝑒 𝑥 𝑑 𝑑𝑥 tanh 𝑥 = 2 𝑒 𝑥 + 𝑒 −𝑥 2 = 𝑠𝑒𝑐ℎ QED

5. WB D2 a) show that 𝑑 𝑑𝑥 arsinh 𝑥 = 1 𝑥 2 +1
b) Find the derivative of 𝑦=𝑎𝑟𝑠𝑖𝑛ℎ (3𝑥+2)
𝑎) 𝑙𝑒𝑡 y=arsinh 𝑥 then 𝑥= sinh 𝑦 and 𝑥 2 = 𝑠𝑖𝑛ℎ 2 𝑦
𝑑𝑥 𝑑𝑦 = cosh 𝑦 = 𝑠𝑖𝑛ℎ 2 𝑦+1
𝑐𝑜𝑠ℎ 2 𝑦− 𝑠𝑖𝑛ℎ 2 𝑦=1
𝑑𝑥 𝑑𝑦 = 𝑥 2 +1
So 𝑑𝑦 𝑑𝑥 = 1 𝑥 QED

6. WB D2 a) show that 𝑑 𝑑𝑥 arsinh 𝑥 = 1 𝑥 2 +1
b) Find the derivative of 𝑦=𝑎𝑟𝑠𝑖𝑛ℎ (3𝑥+2)
𝑏) 𝑑𝑦 𝑑𝑥 = 1 𝑢 ×𝟑
Chain rule
𝒖=3𝑥 y= 𝐚𝐫𝐬𝐢𝐧𝐡 𝒖
𝒖 ′ =𝟑 𝒚 ′ = 1 𝑢 2 +1
𝑏) 𝑑𝑦 𝑑𝑥 = 1 (3𝑥+2) ×𝟑
𝑑𝑦 𝑑𝑥 = 𝑥 2 +12𝑥+5 1/2

7. WB D3 Differentiate with respect to x
a) cosh 3𝑥 b) 𝑥 2 cosh 4𝑥 c) x arcosh 𝑥
sinh 𝑥 ≡ 𝑒 𝑥 − 𝑒 −𝑥 2
cosh 𝑥 ≡ 𝑒 𝑥 + 𝑒 −𝑥 2
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
a) d dx cosh 3𝑥 =3 𝑠𝑖𝑛ℎ 3𝑥
d dx sinh 𝑥 = cosh 𝑥
𝑑 𝑑𝑥 cosh 𝑥 = sinh 𝑥
d dx tanh 𝑥 = 𝑠𝑒𝑐ℎ 2 𝑥
b) 𝑑 𝑑𝑥 𝑥 2 cosh 4𝑥
Product rule
𝒖= 𝒙 𝟐 𝒗= 𝐜𝐨𝐬𝐡 𝟒𝒙
𝒖 ′ =𝟐𝒙 𝒗 ′ =𝟒 𝐬𝐢𝐧𝐡 𝟒𝒙
= 𝑥 2 ×4 sinh 4𝑥 + cosh 4𝑥 ×2𝑥
=4 𝑥 2 sinh 4𝑥 +2𝑥 cosh 4𝑥
d dx arsinh 𝑥 = 1 𝑥 2 +1
𝑑 𝑑𝑥 arcosh 𝑥 = 1 𝑥 2 −1 ,
d dx artanh 𝑥 = 1 1− 𝑥 2
𝑥>1
𝑥 <1
c) d dx x arcosh 𝑥
Product rule
𝒖=𝒙 𝒗= 𝒂𝒓𝐜𝐨𝐬𝐡 𝟒𝒙
𝒖 ′ =𝟏 𝒗 ′ = 1 𝑥 2 −1
=𝑥× 1 𝑥 2 −1 + arcosh 𝑥 ×1
= 𝑥 𝑥 2 −1 + arcosh 𝑥

8. 𝑑 2 𝑦 𝑑 𝑥 2 =9 𝐴 𝑐𝑜𝑠ℎ 3𝑥+𝐵 sinh 3𝑥 =9𝑦 QED
WB D4 Given that 𝑦=𝐴 cosh 3𝑥 +𝐵 sinh 3𝑥 , where A and B are constants
Prove that 𝑑 2 𝑦 𝑑 𝑥 2 =9𝑦
d𝑦 dx =3𝐴𝑠𝑖𝑛ℎ 3𝑥+3𝐵 cosh 3𝑥
𝑑 2 𝑦 𝑑 𝑥 2 =9𝐴 𝑐𝑜𝑠ℎ 3𝑥+9𝐵 sinh 3𝑥
𝑑 2 𝑦 𝑑 𝑥 2 =9 𝐴 𝑐𝑜𝑠ℎ 3𝑥+𝐵 sinh 3𝑥 =9𝑦 QED

9. 𝑥 2 −1 d𝑦 dx 2 =4 arcosh 𝑥 2 =4𝑦 QED d𝑦 dx =2𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 × 1 𝑥 2 −1
WB D Given that 𝑦= arcosh 𝑥 2 ,
Prove that 𝑥 2 −1 𝑑𝑦 𝑑𝑥 2 =4𝑦
d𝑦 dx =2𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 × 1 𝑥 2 −1
Chain rule
𝒖=𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 𝐲= 𝒖 𝟐
𝒖 ′ = 1 𝑥 2 −1 , 𝒚 ′ =2𝑢
Rearrange to
𝑥 2 −1 d𝑦 dx =2𝑎𝑟𝑐𝑜𝑠ℎ 𝑥
Square both sides
𝑥 2 −1 d𝑦 dx 2 =4 arcosh 𝑥 2 =4𝑦 QED

10. 𝑎) 𝑢𝑠𝑒 𝑀𝑎𝑐𝑙𝑎𝑢𝑟𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 then
WB D6
find the first two non-zero terms in the series expansion of 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥
The general term for the series expansion of arsinh x is given by
𝑎𝑟𝑠𝑖𝑛ℎ 𝑥= 𝑟=0 ∞ −1 𝑛 2𝑛 ! 2 2𝑛 𝑛! 𝑥 2𝑛+1 2𝑛+1
b) find , in simplest terms, the terms of the coefficient of 𝑥 5
Use your approximation up to the term in 𝑥 5 to find an approximate value for 𝑎𝑟𝑠𝑖𝑛ℎ 0.5
Calculate the % error for c)
𝑎) 𝑢𝑠𝑒 𝑀𝑎𝑐𝑙𝑎𝑢𝑟𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 then
f(x)=arsinh 𝑥
𝑓(0)=0
𝑓 ′ 𝑥 = 1 𝑥 2 +1
𝑓′(0)=1
𝑓 ′′ 𝑥 =− 𝑥 𝑥 /2
𝑓′′(0)=0
𝑓 ′′′ 𝑥 =− 2 𝑥 2 −1 𝑥 /2
𝑓 ′′′ 0 =−1
𝒂𝒓𝒔𝒊𝒏𝒉 𝒙 =𝒙 − 𝒙 𝟑 𝟔 + …

11. WB D6
find the first two non-zero terms in the series expansion of 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥
The general term for the series expansion of arsinh x is given by
𝑎𝑟𝑠𝑖𝑛ℎ 𝑥= 𝑟=0 ∞ −1 𝑛 2𝑛 ! 2 2𝑛 𝑛! 𝑥 2𝑛+1 2𝑛+1
b) find , in simplest terms, the terms of the coefficient of 𝑥 5
Use your approximation up to the term in 𝑥 5 to find an approximate value for 𝑎𝑟𝑠𝑖𝑛ℎ 0.5
Calculate the % error for c)
𝑔𝑖𝑣𝑒𝑠 − ! ! 𝑥 = 𝑥 5
b) The 𝑥 5 term is when n = 2
𝑎𝑟𝑠𝑖𝑛ℎ 𝑥 =𝑥 − 𝑥 𝑥 …
c) 𝑎𝑟𝑠𝑖𝑛ℎ 0.5 = 0.5 − =
d) % error= −𝑎𝑟𝑠𝑖𝑛ℎ 0.5 𝑎𝑟𝑠𝑖𝑛ℎ 0.5 ×100 = %
𝑎𝑟𝑠𝑖𝑛ℎ 0.5= 𝑒 0.5 − 𝑒 −0.5 2

12. 𝑎) f x =sin x sinh 𝑥 WB D7 𝒚=𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒙 Product rule
Show that 𝑑 4 𝑦 𝑑 𝑥 4 =−4𝑦
Hence find the first three non-zero tems of the Maclaurin series for y, giving each
coefficient in its simplest form
Find an expression for the nth non-zero term of the Maclaurin series for y
𝑎) f x =sin x sinh 𝑥
Product rule
𝑓 ′ 𝑥 = sin 𝑥 cosh 𝑥 + cos 𝑥 sinh 𝑥
Product rule twice
𝑓 ′′ 𝑥 = cos 𝑥 cosh 𝑥 + sin 𝑥 sinh 𝑥 + cos 𝑥 cosh 𝑥 − sin 𝑥 sinh 𝑥
=2 cos 𝑥 cosh 𝑥
Product rule
𝑓 ′′′ 𝑥 =2 cos 𝑥 sinh 𝑥 −2 sin 𝑥 cosh 𝑥
Product rule twice
𝑓 ′𝑣 𝑥 =−2 sin 𝑥 sinh 𝑥 +2 cos 𝑥 cosh 𝑥 −2 sin 𝑥 sinh 𝑥 −2 cos 𝑥 cosh 𝑥
=− 4 sin 𝑥 sinh 𝑥 =− 4𝑦 QED

13. NOW DO EX 6D 𝑏) f x =sin x sinh 𝑥 𝑓(0)=0 𝑓′(0)=0 𝑓′′(0)=2
WB D 𝒚=𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒙
Show that 𝑑 4 𝑦 𝑑 𝑥 4 =−4𝑦
Hence find the first three non-zero tems of the Maclaurin series for y, giving each
coefficient in its simplest form
Find an expression for the nth non-zero term of the Maclaurin series for y
NOW DO EX 6D
𝑏) f x =sin x sinh 𝑥
𝑓 ′ 𝑥 = sin 𝑥 cosh 𝑥 + cos 𝑥 sinh 𝑥
𝑓 ′′′ 𝑥 =2 cos 𝑥 sinh 𝑥 −2 sin 𝑥 cosh 𝑥
𝑓 ′′ 𝑥 =2 cos 𝑥 cosh 𝑥
𝑓 ′𝑣 𝑥 =− 4 sin 𝑥 sinh 𝑥
𝑓(0)=0
𝑓′(0)=0
𝑓′′(0)=2
𝑓 ′′′ 0 =0
𝑓 ′𝑣 0 =0
Find the pattern
𝑓 6 0 =−8
𝑓 6 𝑥 =−8 cos 𝑥 cosh 𝑥
Q From exam materials 2018
𝑓 10 𝑥 =32 cos 𝑥 cosh 𝑥
𝑓 10 (0)=32
𝒙 𝟐 𝟐 𝟐 + 𝒙 𝟔 𝟔 ! −𝟖 + 𝒙 𝟏𝟎 𝟏𝟎 ! 𝟑𝟐 +…
𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒙 =
= 𝒙 𝟐 − 𝒙 𝟔 𝟗𝟎 + 𝒙 𝟏𝟎 𝟏𝟏𝟑𝟒𝟎𝟎 +…
𝒄) 𝒙 𝟒𝒏−𝟐 𝟒𝒏−𝟐 ! 𝟐 −𝟒 𝒏−𝟏

14. One thing to improve is –
KUS objectives
BAT differentiate and integrate with hyperbolic functions
self-assess
One thing learned is –
One thing to improve is –

15. END