Hyperbolic functions

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 = 1 2 𝑠𝑒𝑐 2 𝑥 2 tan 𝑥 2

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

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) 1 2 𝑒 −𝑥 = 1 2 𝑒 𝑥 + 1 2 𝑒 −𝑥 = cosh 𝑥 𝑏) cosh 𝑥 ≡ 𝑒 𝑥 + 𝑒 −𝑥 2 = 𝑒 𝑥 2 + 𝑒 −𝑥 2 𝑑 𝑑𝑥 sinh 𝑥 = 1 2 𝑒 𝑥 +(−1) 1 2 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 +1 2𝑒 2𝑥 − 𝑒 2𝑥 −1 2𝑒 2𝑥 𝑒 2𝑥 +1 2 𝑑 𝑑𝑥 tanh 𝑥 = 2𝑒 4𝑥 + 2𝑒 2𝑥 − 2𝑒 4𝑥 + 2𝑒 2𝑥 𝑒 2𝑥 +1 2 = 4𝑒 2𝑥 𝑒 2𝑥 +1 2 = 2𝑒 𝑥 2 𝑒 2𝑥 +1 2 = 2𝑒 𝑥 𝑒 2𝑥 +1 2 divide through by 𝑒 𝑥 𝑑 𝑑𝑥 tanh 𝑥 = 2 𝑒 𝑥 + 𝑒 −𝑥 2 = 𝑠𝑒𝑐ℎ 2 QED

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 2 +1 QED

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 2 +1 ×𝟑 Chain rule 𝒖=3𝑥+2 y= 𝐚𝐫𝐬𝐢𝐧𝐡 𝒖 𝒖 ′ =𝟑 𝒚 ′ = 1 𝑢 2 +1 𝑏) 𝑑𝑦 𝑑𝑥 = 1 (3𝑥+2) 2 +1 ×𝟑 𝑑𝑦 𝑑𝑥 = 3 9 𝑥 2 +12𝑥+5 1/2

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

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

𝑥 2 −1 d𝑦 dx 2 =4 arcosh 𝑥 2 =4𝑦 QED d𝑦 dx =2𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 × 1 𝑥 2 −1 WB D5 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

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! 2 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 +1 3/2 𝑓′′(0)=0 𝑓 ′′′ 𝑥 =− 2 𝑥 2 −1 𝑥 2 +1 5/2 𝑓 ′′′ 0 =−1 𝒂𝒓𝒔𝒊𝒏𝒉 𝒙 =𝒙 − 𝒙 𝟑 𝟔 + …

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! 2 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 −1 2 4 ! 2 4 2! 2 𝑥 5 5 = 3 40 𝑥 5 b) The 𝑥 5 term is when n = 2 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥 =𝑥 − 𝑥 3 6 + 3 𝑥 5 40 … c) 𝑎𝑟𝑠𝑖𝑛ℎ 0.5 = 0.5 − 0.5 3 6 + 3 0.5 5 40 =0.48151 d) % error= 0.48151 −𝑎𝑟𝑠𝑖𝑛ℎ 0.5 𝑎𝑟𝑠𝑖𝑛ℎ 0.5 ×100 = 0.062% 𝑎𝑟𝑠𝑖𝑛ℎ 0.5= 𝑒 0.5 − 𝑒 −0.5 2

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

NOW DO EX 6D 𝑏) f x =sin x sinh 𝑥 𝑓(0)=0 𝑓′(0)=0 𝑓′′(0)=2 WB D7 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

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 –

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