1. FP3 Chapter 3 Differentiation
Dr J Frost
Last modified: 5th December 2016

2. Overview
This (very short!) chapter builds on the previous ones regarding hyperbolic functions.
We’ll see how to differentiate hyperbolic and inverse hyperbolic functions.
We’ll also see the derivative of inverse trigonometric functions (but you can be asked to prove these in C3 as well!)

3. Differentiating hyperbolic functions
𝑑 𝑑𝑥 sinh 𝑥 = cosh 𝑥 𝑑 𝑑𝑥 cosh 𝑥 = sinh 𝑥 𝑑 𝑑𝑥 tanh 𝑥 = sech 2 𝑥 𝑑 𝑑𝑥 coth 𝑥 =−𝑐𝑜𝑠𝑒𝑐 ℎ 2 𝑥
Important Bro Memorisation Tip: They’re all the same as non-hyperbolic results, other than that 𝑐𝑜𝑠ℎ is not negated and sech 𝑥 becomes − sech 𝑥 tanℎ 𝑥 (i.e. is negated).
Prove that 𝑑 𝑑𝑥 sinh 𝑥 = cosh 𝑥 ?
sinh 𝑥 = 𝑒 𝑥 − 𝑒 −𝑥 2 𝑑 𝑑𝑥 sinh 𝑥 = 𝑒 𝑥 + 𝑒 −𝑥 2 = cosh 𝑥

4. Test Your Understanding
Hint: Did someone say chain rule? ?

5. Exercise 3A
In Q1-16, differentiate with respect to 𝑥.
If 𝑦=𝑎 cosh 𝑛𝑥 +𝑏 sinh 𝑛𝑥 , where 𝑎 and 𝑏 are constants, prove that 𝑑 2 𝑦 𝑑 𝑥 2 = 𝑛 2 𝑦.
Find the stationary values of the curve with equation 𝑦=12 cosh 𝑥 − sinh 𝑥 .
𝐭𝐚𝐧𝐡 −𝟏 𝟏 𝟏𝟐
Given that 𝑦= cosh 3𝑥 sinh 𝑥 , find 𝑑 2 𝑦 𝑑 𝑥 2 .
𝟐 𝟓 𝐜𝐨𝐬𝐡 𝟑𝒙 𝐬𝐢𝐧𝐡 𝒙 +𝟑 𝐬𝐢𝐧𝐡 𝟑𝒙 𝐜𝐨𝐬𝐡 𝒙
Find the equation of the tangent and normal to the hyperbola 𝑥 − 𝑦 =1 17
sinh 2𝑥 → 2 cosh 2𝑥 cosh 5𝑥 → 5 sinh 5𝑥 tanh 2𝑥 →2 sech 2 2𝑥 sinh 3𝑥 →3 cosh 3𝑥 coth 4𝑥 →−4𝑐𝑜𝑠𝑒𝑐 ℎ 2 4𝑥 sech 2𝑥 →−2 tanh 2𝑥 sech 2𝑥 𝑒 −𝑥 sinh 𝑥 → 𝑒 −𝑥 cosh 𝑥 − sinh 𝑥 𝑥 cosh 3𝑥 → cosh 3𝑥 +3 sinh 3𝑥 sinh 𝑥 3𝑥 → 𝑥 cosh 𝑥 − sinh 𝑥 3 𝑥 2 𝑥 2 cosh 3𝑥 → 𝑥(2 cosh 3𝑥 +3𝑥 sinh 3𝑥 ) sinh 2𝑥 cosh 3𝑥 → 2 cosh 2𝑥 cosh 3𝑥 sinh 2𝑥 sinh 3𝑥 ln cosh 𝑥 → tanh 𝑥 sinh 𝑥 →3 𝑥 2 cosh 𝑥 cosh 2 2𝑥 →4 cosh 2𝑥 sinh 2𝑥 𝑒 cosh 𝑥 → sinh 𝑥 𝑒 cosh 𝑥 𝑐𝑜𝑠𝑒𝑐ℎ 𝑥 → − coth 𝑥 𝑐𝑜𝑠𝑒𝑐 𝑥 ? 1 2 ? 3 ? 4 ? 18 5 ? ? 6 ? 7 ? 8 ? ? 19 9 ? 10 ? 11 ? 20 12 ? 13 ? 14 ? 15 ? 16 ?

6. Inverse Hyperbolic Functions
Proof
𝑑 𝑑𝑥 arsinh 𝑥 = 1 𝑥 𝑑 𝑑𝑥 arcosh 𝑥 = 1 𝑥 2 −1 𝑑 𝑑𝑥 artanh 𝑥 = 1 1− 𝑥 2
𝑦=𝑎𝑟𝑠𝑖𝑛ℎ 𝑥 sinh 𝑦 =𝑥 𝑑𝑥 𝑑𝑦 = cosh 𝑦 𝑑𝑥 𝑑𝑦 = sinh 2 𝑦 = 𝑥 𝑑𝑦 𝑑𝑥 = 1 𝑥 2 +1 ? ? ? ?
Examples
Given that 𝑦= 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 2 prove that 𝑥 2 −1 𝑑𝑦 𝑑𝑥 2 =4𝑦
Find 𝑑 𝑑𝑥 𝑎𝑟𝑡𝑎𝑛ℎ 3𝑥
By chain rule:
𝒅 𝒅𝒙 𝒂𝒓𝒕𝒂𝒏𝒉 𝟑𝒙 = 𝟏 𝟏− 𝟑𝒙 𝟐 ×𝟑 = 𝟑 𝟏−𝟗 𝒙 𝟐
𝒅𝒚 𝒅𝒙 =𝟐 𝒂𝒓𝒄𝒐𝒔𝒉 𝒙 𝟏 𝒙 𝟐 −𝟏 𝒙 𝟐 −𝟏 𝒅𝒚 𝒅𝒙 =𝟐 𝒂𝒓𝒄𝒐𝒔𝒉 𝒙 𝒙 𝟐 −𝟏 𝒅𝒚 𝒅𝒙 𝟐 =𝟒 𝒂𝒓𝒄𝒐𝒔𝒉 𝒙 𝟐 𝒙 𝟐 −𝟏 𝒅𝒚 𝒅𝒙 𝟐 =𝟒𝒚 ? ?

7. Test Yarrr Understanding? ? ?

8. Exercise 3B
Differentiate
𝑎𝑟𝑐𝑜𝑠ℎ 2𝑥 → 𝑥 2 −1 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥+1 → 𝑥 𝑎𝑟𝑡𝑎𝑛ℎ 3𝑥 → 3 1−9 𝑥 2 𝑎𝑟𝑠𝑒𝑐ℎ 𝑥 →− 1 𝑥 1− 𝑥 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 → 2𝑥 𝑥 4 −1 𝑎𝑟𝑐𝑜𝑠ℎ 3𝑥 → 𝑥 2 −1 𝑥 2 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 →2𝑥 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥+ 𝑥 𝑥 2 −1 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥 → 1 𝑥 𝑒 𝑥 3 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥 →3 𝑥 2 𝑒 𝑥 3 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥+ 𝑒 𝑥 𝑥 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 → 1 𝑥 sech 𝑥 − 1 𝑥 2 −1 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 sech 𝑥 → sech 𝑥 𝑥 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥𝑡𝑎𝑛ℎ 𝑥 𝑥 𝑎𝑟𝑐𝑜𝑠ℎ 3𝑥 →𝑎𝑟𝑐𝑜𝑠ℎ 3𝑥+ 3𝑥 9 𝑥 2 −1
Prove that
𝑑 𝑑𝑥 cosh 𝑥 = 1 𝑥 2 −1 𝑑 𝑑𝑥 𝑎𝑟𝑡𝑎𝑛ℎ 𝑥 = 1 1− 𝑥 2
Given that 𝑦=𝑎𝑟𝑡𝑎𝑛ℎ 𝑒 𝑥 2 , prove that
4− 𝑒 2𝑥 𝑑𝑦 𝑑𝑥 =2 𝑒 𝑥
Given that 𝑦=𝑎𝑟𝑠𝑖𝑛ℎ 𝑥, show that
1+ 𝑥 2 𝑑 3 𝑦 𝑑 𝑥 3 +3𝑥 𝑑 2 𝑦 𝑑 𝑥 2 + 𝑑𝑦 𝑑𝑥 =0
If 𝑦= 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 2 , find 𝑑 2 𝑦 𝑑 𝑥 2
−𝟐𝒙 𝒂𝒓𝒄𝒐𝒔𝒉 𝒙 𝒙 𝟐 −𝟏 𝟑 𝟐
Find the equation of the tangent at the point where 𝑥= on the curve with equation
𝑦=𝑎𝑟𝑡𝑎𝑛ℎ 𝑥.
𝟐𝟓𝒚−𝟐𝟓 𝐥𝐧 𝟓 =𝟏𝟔𝟗𝒙−𝟏𝟓𝟔 1 2 ? a ? b ? c ? d 3 ? e ? f 4 ? g ? h 5 ? i ? ? j 6 ? k ? l ?

9. Inverse Trig Functions
You can actually be asked to prove these in C4 examinations!
Proof
𝑦= arcsin 𝑥 sin 𝑦 =𝑥 cos 𝑦 𝑑𝑦 𝑑𝑥 =1 𝑑𝑦 𝑑𝑥 = 1 cos 𝑦 𝑑𝑦 𝑑𝑥 = 1 1− sin 2 𝑥 𝑑𝑦 𝑑𝑥 = 1 1− 𝑥 2 ?
𝑑 𝑑𝑥 arcsin 𝑥 = 1 1− 𝑥 𝑑 𝑑𝑥 arccos 𝑥 =− 1 1− 𝑥 𝑑 𝑑𝑥 arctan 𝑥 = 1 1+ 𝑥 2 ? ?
Bro Exam Note: All 9 results in this chapter are in the formula booklet!

10. Inverse Trig Functions
Given 𝑦= arcsin 𝑥 2 , find 𝑑𝑦 𝑑𝑥
𝑑 𝑑𝑥 arcsin 𝑥 = 1 1− 𝑥 𝑑 𝑑𝑥 arccos 𝑥 =− 1 1− 𝑥 𝑑 𝑑𝑥 arctan 𝑥 = 1 1+ 𝑥 2
Using chain rule:
𝑑𝑦 𝑑𝑥 = 1 1− 𝑥 4 ×2𝑥= 2𝑥 1− 𝑥 4 ?
Given 𝑦= tan −1 1−𝑥 1+𝑥 , find 𝑑𝑦 𝑑𝑥
Either using chain rule with standard result, or do from scratch:
tan 𝑦 = 1−𝑥 1+𝑥 sec 2 𝑦 𝑑𝑦 𝑑𝑥 = − 1+𝑥 −1 1−𝑥 1+𝑥 𝑑𝑦 𝑑𝑥 = 1 sec 2 𝑦 ×− 𝑥 𝑑𝑦 𝑑𝑥 = tan 2 𝑦 ×− 𝑥 2 ?
= −𝑥 1+𝑥 2 ×− 𝑥 =… =− 1 1+ 𝑥 2

11. Test Your Understanding?

12. Exercise 3C
𝑑 𝑑𝑥 𝑒 𝑥 arccos 𝑥 = 𝒆 𝒙 𝐚𝐫𝐜𝐜𝐨𝐬 𝒙 − 𝟏 𝟏− 𝒙 𝟐 𝑑 𝑑𝑥 arcsin 𝑥 cos 𝑥 = 𝐜𝐨𝐬 𝒙 𝟏− 𝒙 𝟐 − 𝐬𝐢𝐧 𝒙 𝐚𝐫𝐜𝐬𝐢𝐧 𝒙 𝑑 𝑑𝑥 𝑥 2 arccos 𝑥 =𝒙 𝟐 𝒂𝒓𝒄𝒄𝒐𝒔 𝒙 − 𝒙 𝟏− 𝒙 𝟐 𝑑 𝑑𝑥 𝑒 arctan 𝑥 = 𝒆 𝒂𝒓𝒄𝒕𝒂𝒏 𝒙 𝟏+ 𝒙 𝟐
Given that 𝑦= arccos 𝑥 , prove that 𝑑𝑦 𝑑𝑥 =− 1 1− 𝑥 2
Differentiate with respect to 𝑥.
𝑑 𝑑𝑥 arccos 2𝑥 =− 𝟐 𝟏−𝟒 𝒙 𝟐 𝑑 𝑑𝑥 arctan 𝑥 2 = 𝟐 𝒙 𝟐 +𝟒 𝑑 𝑑𝑥 arcsin 3𝑥 = 𝟑 𝟏−𝟗 𝒙 𝟐 𝑑 𝑑𝑥 arccot 𝑥 =− 𝟏 𝟏+ 𝒙 𝟐 𝑑 𝑑𝑥 arcsec 𝑥 = 𝟏 𝒙 𝒙 𝟐 −𝟏 𝑑 𝑑𝑥 𝑎𝑟𝑐𝑐𝑜𝑠𝑒𝑐 𝑥 =− 𝟏 𝒙 𝒙 𝟐 −𝟏 𝑑 𝑑𝑥 arcsin 𝑥 𝑥−1 =− 𝟏 𝒙−𝟏 𝟏−𝟐𝒙 𝑑 𝑑𝑥 arccos 𝑥 2 =− 𝟐𝒙 𝟏− 𝒙 𝟒 ? 1 i ? j 2 ? ? k a ? ? l b ? c
If tan 𝑦 =𝑥 arctan 𝑥 , find 𝑑𝑦 𝑑𝑥
𝟏 𝟏+ 𝒙 𝟐 𝐚𝐫𝐜𝐭𝐚𝐧 𝒙 𝟐 𝐚𝐫𝐜𝐭𝐚𝐧 𝒙 + 𝒙 𝟏+ 𝒙 𝟐
Given that 𝑦= arcsin 𝑥 , prove that 1− 𝑥 2 𝑑 2 𝑦 𝑑 𝑥 2 −𝑥 𝑑𝑦 𝑑𝑥 =0
Find an equation of the tangent to the curve with equation 𝑦= arcsin 2𝑥 at the point 𝑥= 𝟑 𝒚− 𝝅 𝟑 𝟔 =𝟒𝒙−𝟏 3 ? ? d ? e 4 ? f ? g 5 ? h ?