1. Differentiation

2. Introduction
Up to this point you have seen how to differentiate expressions
You have also seen that differentiating gives the gradient function
You have seen that differentiating again indicates whether turning points are maxima or minima
In this Chapter you will learn how to differentiate much more complicated expressions
These will include functions of functions, exponentials (e), logarithms and trigonometric graphs

3. Teachings for Exercise 8A

4. Differentiation 𝑑𝑦 𝑑𝑥 =𝑛[𝑓 𝑥 ] 𝑛−1 𝑓′(𝑥) 𝑦=[𝑓 𝑥 ] 𝑛
You need to be able to differentiate a function of a function, using the Chain Rule
Differentiate:
𝑦=(3 𝑥 4 +𝑥 ) 5
𝑑𝑦 𝑑𝑥 =𝑛[𝑓 𝑥 ] 𝑛−1 𝑓′(𝑥)
The whole function is raised to a power
𝑦=[𝑓 𝑥 ] 𝑛
𝑑𝑦 𝑑𝑥 =𝑛[𝑓 𝑥 ] 𝑛−1 𝑓′(𝑥)
Multiply by the power and reduce it by 1 ‘as usual’
Also multiply by f’(x)
𝑓(𝑥)=3 𝑥 4 +𝑥
Differentiate f(x) to get f’(x)
𝑓′(𝑥)=12 𝑥 3 +1
n is just the power on the bracket
𝑛=5
𝑑𝑦 𝑑𝑥 =5
(3 𝑥 4 +𝑥 ) 4
(12 𝑥 3 +1) 8A

5. Differentiation 𝑑𝑦 𝑑𝑥 =𝑛[𝑓 𝑥 ] 𝑛−1 𝑓′(𝑥) 𝑦=[𝑓 𝑥 ] 𝑛
𝑦= 5 𝑥 2 +1
You need to be able to differentiate a function of a function, using the Chain Rule
Rewrite in a ‘differentiatable’ form
𝑦=(5 𝑥 2 +1 ) 1 2
𝑑𝑦 𝑑𝑥 =𝑛[𝑓 𝑥 ] 𝑛−1 𝑓′(𝑥)
The whole function is raised to a power
𝑦=[𝑓 𝑥 ] 𝑛
Multiply by the power and reduce it by 1 ‘as usual’
Also multiply by f’(x)
𝑑𝑦 𝑑𝑥 =𝑛[𝑓 𝑥 ] 𝑛−1 𝑓′(𝑥)
𝑓(𝑥)=5 𝑥 2 +1
Differentiate f(x)
𝑓′(𝑥)=10𝑥
n is just the power
𝑛= 1 2
𝑦= 5 𝑥 2 +1
Given that:
𝑑𝑦 𝑑𝑥 =
Find the value of f’(x) at (4,9)
1 2
(5 𝑥 2 +1 ) − 1 2
(10𝑥)
The ½ and 10x can be combined
𝑑𝑦 𝑑𝑥 =
(5 𝑥 2 +1 ) − 1 2 5𝑥
Sub in 4 from the coordinate (calculator ok!)
=2 2 9 8A

6. Differentiation 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 × 𝑑𝑢 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 × 𝑑𝑢 𝑑𝑥 𝑑𝑦 𝑑𝑥 =
You need to be able to differentiate a function of a function, using the Chain Rule
𝑦=( 𝑥 2 −7𝑥 ) 4
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 × 𝑑𝑢 𝑑𝑥
This is another form of the Chain Rule
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 × 𝑑𝑢 𝑑𝑥
y is a function of u, and u is a function of x
𝑢= 𝑥 2 −7𝑥
Let u be the part in brackets, and differentiate
𝑦= 𝑢 4
Differentiate y in terms of u
𝑑𝑢 𝑑𝑥 =2𝑥−7
𝑑𝑦 𝑑𝑢 =4 𝑢 3
Given that:
𝑦=( 𝑥 2 −7𝑥 ) 4
𝑑𝑦 𝑑𝑥 =
4 𝑢 3
(2𝑥−7)
𝑑𝑦 𝑑𝑥
Calculate
using the chain rule
Sub in u = x2 – 7x
𝑑𝑦 𝑑𝑥 =
4 ( 𝑥 2 −7𝑥) 3
(2𝑥−7) 8A

7. Differentiation 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 × 𝑑𝑢 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 × 𝑑𝑢 𝑑𝑥 𝑑𝑦 𝑑𝑥 =
𝑦= 1 6𝑥−3
You need to be able to differentiate a function of a function, using the Chain Rule
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 × 𝑑𝑢 𝑑𝑥
This is another form of the Chain Rule
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 × 𝑑𝑢 𝑑𝑥
𝑦= 𝑢 − 1 2
y is a function of u, and u is a function of x
Let u be the part under the square root, and differentiate
𝑢=6𝑥−3
Differentiate in terms of u
𝑑𝑢 𝑑𝑥 =6
𝑑𝑦 𝑑𝑢 =− 1 2 𝑢 − 3 2
𝑦= 1 6𝑥−3
Given that:
𝑑𝑦 𝑑𝑥 =
− 1 2 𝑢 − 3 2
𝑑𝑦 𝑑𝑥
(6)
Calculate
using the chain rule
Combine the 6 and -1/2
𝑑𝑦 𝑑𝑥 =
−3 𝑢 − 3 2
Substitute in u
𝑑𝑦 𝑑𝑥 =
−3 (6𝑥−3) − 3 2 8A

8. Differentiation 𝑑𝑦 𝑑𝑥 = 1 𝑑𝑥 𝑑𝑦 𝑑𝑦 𝑑𝑥 8A 𝑥=𝑦 3 +𝑦 𝑑𝑥 𝑑𝑦 =3𝑦 2 +1
You need to be able to differentiate a function of a function, using the Chain Rule
Differentiate, only ‘the other way round’
(no real difference…)
𝑑𝑥 𝑑𝑦 =3𝑦 2 +1
Invert the answer to get dy/dx
𝑑𝑦 𝑑𝑥 = 1 𝑑𝑥 𝑑𝑦
This is a particular case of differentiating…
𝑑𝑦 𝑑𝑥 =
1 3 𝑦 2 +1
Sub in the y-coordinate since the equation is in terms of y (you’re used to only using x!)
1 3 (1) 2 +1 =
𝑑𝑦 𝑑𝑥
Calculate
for the following equation,
And work it out!
= 1 4
and find the gradient at the point (2,1),
𝑦 3 +𝑦=𝑥 8A

9. Teachings for Exercise 8B

10. Differentiation 8B Let y = uv, where u and v are both functions of x
You need to be able to differentiate functions that are multiplied together, using the product rule
𝑦=𝑢𝑣
A small change in x, would cause a small change in u and v, and consequently also in y
δ = small change
If: 𝑦=𝑢𝑣
𝑦+δ𝑦=(𝑢+δ𝑢)(𝑣+δ𝑣)
y = uv
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
𝑢𝑣+δ𝑦=(𝑢+δ𝑢)(𝑣+δ𝑣)
Subtract uv
δ𝑦= 𝑢+δ𝑢 𝑣+δ𝑣 −𝑢𝑣
The Product Rule
Multiply out the brackets
You do not need to learn the proof, but you do need to be able to use this result!
δ𝑦=𝑢𝑣+𝑢δ𝑣+𝑣δ𝑢+δ𝑢δ𝑣−𝑢𝑣
Cancel uv and -uv
δ𝑦=𝑢δ𝑣+𝑣δ𝑢+δ𝑢δ𝑣
Divide all by δx
δ𝑦 δ𝑥 =𝑢 δ𝑣 δ𝑥 +𝑣 δ𝑢 δ𝑥 +δ𝑣 δ𝑢 δ𝑥
As δx  0, δy/δx = dy/dx and so on….
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥 +δ𝑣 𝑑𝑢 𝑑𝑥
As δx  0, δy, δu and δv also  0
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥 8B

11. Differentiation 8B If: 𝑦=𝑢𝑣 𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
Given that: 𝑓 𝑥 =𝑥(2𝑥+1 ) 3 , find f’(x)
𝑢=𝑥
𝑣=(2𝑥+1 ) 3
𝑓 𝑥 =𝑥(2𝑥+1 ) 3
Simple
Use the chain rule from 8A
𝑑𝑢 𝑑𝑥 =1
𝑑𝑣 𝑑𝑥 =3 (2𝑥+1 ) 2 (2)
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
Simplify if possible
𝑑𝑣 𝑑𝑥 =6(2𝑥+1 ) 2
𝑑𝑦 𝑑𝑥 = 𝑥
6(2𝑥+1 ) 2
+ (2𝑥+1 ) 3 1
Try to rewrite some parts or group terms
𝑑𝑦 𝑑𝑥 =
6𝑥(2𝑥+1 ) 2
+ (2𝑥+1 ) 3
(2x + 1)2 is a common factor to both terms, so factorise
𝑑𝑦 𝑑𝑥 =
(2𝑥+1 ) 2
(6𝑥+ 2𝑥+1 )
You can also now simplify the part in brackets
𝑑𝑦 𝑑𝑥 =
(2𝑥+1 ) 2
(8𝑥+1) 8B

12. Differentiation 8B If: 𝑦=𝑢𝑣 𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
Given that: 𝑓 𝑥 = 𝑥 2 3𝑥−1 , find f’(x)
𝑣=(3𝑥−1 ) 1 2
𝑓 𝑥 = 𝑥 2 3𝑥−1
𝑢= 𝑥 2
Use the chain rule from 8A
Simple
𝑑𝑣 𝑑𝑥 = 1 2 (3𝑥−1 ) − 1 2 (3)
𝑑𝑢 𝑑𝑥 =2𝑥
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
Simplify if possible
𝑑𝑣 𝑑𝑥 = 3 2 (3𝑥−1 ) − 1 2
𝑑𝑦 𝑑𝑥 =
3 2 (3𝑥−1 ) − 1 2
+ (3𝑥−1 ) 1 2
𝑥 2 2𝑥
Try to rewrite some parts or group terms
𝑑𝑦 𝑑𝑥 =
3 𝑥 2 2 (3𝑥−1 ) − 1 2
+ 2𝑥(3𝑥−1 ) 1 2
A Power ½ = √x
A Negative Power = 1/x
𝑑𝑦 𝑑𝑥 =
3 𝑥 𝑥−1
+ 2𝑥 3𝑥−1
Multiply the right by 2√(3x-1) (Top and ‘Bottom’) so the denominators are the same
𝑑𝑦 𝑑𝑥 =
3 𝑥 𝑥−1
+ 4𝑥 3𝑥−1 3𝑥− 𝑥−1
You have a ‘square root squared’ on the right
𝑑𝑦 𝑑𝑥 =
3 𝑥 𝑥−1
+ 4𝑥(3𝑥−1) 2 3𝑥−1
𝑑𝑦 𝑑𝑥 =
15 𝑥 2 −4𝑥 2 3𝑥−1
Group up like terms 8B

13. Teachings for Exercise 8C

14. You need to be able to differentiate functions using the quotient rule
Differentiation
You need to be able to differentiate functions using the quotient rule
If: 𝑦= 𝑢 𝑣 where u and v are functions of x…
𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2
The Quotient Rule 8C

15. Differentiation 8C 𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2 𝑦= 𝑥 2𝑥+5 𝑢=𝑥 𝑣=2𝑥+5
𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2
Differentiation
𝑦= 𝑥 2𝑥+5
𝑢=𝑥
𝑣=2𝑥+5
Given that:
Differentiate
Differentiate
𝑑𝑢 𝑑𝑥 =1
𝑑𝑣 𝑑𝑥 =2
Calculate 𝑑𝑦 𝑑𝑥
𝑦= 𝑥 2𝑥+5
𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2
Substitute in each component
𝑑𝑦 𝑑𝑥 =
(2𝑥+5)(1)
− (𝑥)(2)
(2𝑥+5 ) 2
Simplify/work out parts of the numerator
𝑑𝑦 𝑑𝑥 =
2𝑥+5
− 2𝑥
(2𝑥+5 ) 2
Group together like terms
𝑑𝑦 𝑑𝑥 = 5
(2𝑥+5 ) 2 8C

16. Differentiation 8C 𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2 𝑦= 3𝑥 (𝑥+4 ) 3 𝑢=3𝑥
𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2
Differentiation
𝑦= 3𝑥 (𝑥+4 ) 3
𝑢=3𝑥
𝑣=(𝑥+4 ) 3
Given that:
Differentiate using the chain rule
Differentiate
𝑑𝑢 𝑑𝑥 =3
𝑑𝑣 𝑑𝑥 =3(𝑥+4 ) 2 (1)
Calculate 𝑑𝑦 𝑑𝑥
Simplify if possible
𝑦= 3𝑥 (𝑥+4 ) 3
𝑑𝑣 𝑑𝑥 =3(𝑥+4 ) 2
𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2
Substitute in each component
𝑑𝑦 𝑑𝑥 =
(𝑥+4 ) 3 (3)
− (3𝑥)(3(𝑥+4 ) 2 )
((𝑥+4 ) 3 ) 2
Simplify/work out parts of the numerator 1
𝑑𝑦 𝑑𝑥 =
3(𝑥+4 ) 3
− 9𝑥(𝑥+4 ) 2
(𝑥+4 ) 6 4
Cancel out (x + 4)2
𝑑𝑦 𝑑𝑥 =
3(𝑥+4)
− 9𝑥
(𝑥+4 ) 4
Multiply out and re-factorise the numerator
𝑑𝑦 𝑑𝑥 =
6(2−𝑥)
(𝑥+4 ) 4 8C

17. Teachings for Exercise 8D

18. You need to be able to differentiate the exponential function
Differentiation
Differentiate the following:
You need to be able to differentiate the exponential function a)
𝑦=5 𝑒 𝑥
Having a number in front does not affect dy/dx
If:
𝑦= 𝑒 𝑥
𝑑𝑦 𝑑𝑥 =
5 𝑒 𝑥
𝑑𝑦 𝑑𝑥 = 𝑒 𝑥
Then: b)
𝑦= 𝑒 2𝑥+3
If:
𝑦= 𝑒 𝑓(𝑥)
Differentiate the power and put it in front
𝑑𝑦 𝑑𝑥 =
2 𝑒 2𝑥+3
𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥)𝑒 𝑓(𝑥)
Then:
Differentiate the power, and put it in front of ef(x) 8D

19. Differentiation 8D If: 𝑦= 𝑒 𝑓(𝑥) 𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥)𝑒 𝑓(𝑥) Then:
Differentiate the following:
𝑢=𝑥
𝑣= 𝑒 𝑥 2
Differentiate the power and put it in front
Differentiate
𝑦=𝑥 𝑒 𝑥 2
Use the product rule as we have 2 functions multiplied
𝑑𝑢 𝑑𝑥 =1
𝑑𝑣 𝑑𝑥 = 2𝑥𝑒 𝑥 2
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
Substitute the values in
𝑑𝑦 𝑑𝑥 =
(𝑥)
(2𝑥 𝑒 𝑥 2 )
+ ( 𝑒 𝑥 2 )
(1)
Simplify some parts if possible
𝑑𝑦 𝑑𝑥 =
2 𝑥 2 𝑒 𝑥 2
+ 𝑒 𝑥 2
Factorise with 𝑒 𝑥 2 as the common factor
𝑑𝑦 𝑑𝑥 =
𝑒 𝑥 2
(2 𝑥 2 +1) 8D

20. Differentiation 8D If: 𝑦= 𝑒 𝑓(𝑥) 𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥)𝑒 𝑓(𝑥) Then:
Differentiate the following:
𝑢= 𝑒 3𝑥−1
𝑣=𝑥
Differentiate the power and put it in front
Differentiate
𝑦= 𝑒 3𝑥−1 𝑥
𝑑𝑢 𝑑𝑥 =3 𝑒 3𝑥−1
𝑑𝑣 𝑑𝑥 =1
Use the quotient as we have 2 functions in a division
𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2
Substitute the values in
𝑑𝑦 𝑑𝑥 =
(𝑥)
(3 𝑒 3𝑥−1 )
− ( 𝑒 3𝑥−1 )
(1)
𝑥 2
Simplify some parts if possible
𝑑𝑦 𝑑𝑥 =
3𝑥 𝑒 3𝑥−1
− 𝑒 3𝑥−1
𝑥 2
Factorise with 𝑒 3𝑥−1 as the common factor
𝑑𝑦 𝑑𝑥 =
𝑒 3𝑥−1 (3𝑥−1)
𝑥 2 8D

21. Teachings for Exercise 8E

22. You need to be able to differentiate the logarithmic function
Differentiation
You need to be able to differentiate the logarithmic function
If:
𝑦=𝑙𝑛𝑥
𝑑𝑦 𝑑𝑥 = 1 𝑥
Then:
Let:
𝑦=𝑙𝑛𝑥
Inverse Logarithm
𝑥= 𝑒 𝑦
Differentiate
𝑑𝑥 𝑑𝑦 = 𝑒 𝑦
If:
𝑦=ln⁡[𝑓 𝑥 ]
Reverse Differentiation
𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥) 𝑓(𝑥)
Then:
𝑑𝑦 𝑑𝑥 = 1 𝑒 𝑦
Earlier we said x = ey
𝑑𝑦 𝑑𝑥 = 1 𝑥 8E

23. You need to be able to differentiate the logarithmic function
Differentiation
Differentiate:
You need to be able to differentiate the logarithmic function a)
𝑦=5𝑙𝑛𝑥
Differentiate
𝑑𝑦 𝑑𝑥 = 5 𝑥
If:
𝑦=𝑙𝑛𝑥
𝑑𝑦 𝑑𝑥 = 1 𝑥
Then: b)
𝑦=ln⁡(6𝑥−1)
Differentiating ln of a function
If:
𝑦=ln⁡[𝑓 𝑥 ]
𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥) 𝑓(𝑥)
You can probably do this in your head!
𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥) 𝑓(𝑥)
Then:
𝑑𝑦 𝑑𝑥 = 6 6𝑥−1 8E

24. Factorise (non-essential)
If:
𝑦=𝑙𝑛𝑥
If:
𝑦=ln⁡[𝑓 𝑥 ]
Differentiation
𝑑𝑦 𝑑𝑥 = 1 𝑥
𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥) 𝑓(𝑥)
Then:
Then:
Differentiate the following:
𝑢= 𝑥 3
𝑣=𝑙𝑛𝑥 c)
𝑦= 𝑥 3 𝑙𝑛𝑥
Differentiate
Differentiate
Use the product rule
𝑑𝑢 𝑑𝑥 = 3𝑥 2
𝑑𝑣 𝑑𝑥 = 1 𝑥
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
Sub in values
𝑑𝑦 𝑑𝑥 =
1 𝑥
(𝑥 3 )
+ (𝑙𝑛𝑥)
(3 𝑥 2 )
Simplify terms
𝑑𝑦 𝑑𝑥 =
𝑥 2
+ 3𝑥 2 𝑙𝑛𝑥
Factorise (non-essential)
𝑑𝑦 𝑑𝑥 =
𝑥 2 (1+3𝑙𝑛𝑥) 8E

25. Differentiation 8E If: 𝑦=𝑙𝑛𝑥 If: 𝑦=ln⁡[𝑓 𝑥 ] 𝑑𝑦 𝑑𝑥 = 1 𝑥
𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥) 𝑓(𝑥)
Then:
Then:
Differentiate the following:
𝑢=𝑙𝑛5𝑥
𝑣=𝑥
𝑦= 𝑙𝑛5𝑥 𝑥 d)
Differentiate
Differentiate
𝑑𝑢 𝑑𝑥 = 5 5𝑥
𝑑𝑣 𝑑𝑥 =1
Use the quotient rule
𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2
Sub in values
5 5𝑥
(𝑥)
− 𝑙𝑛5𝑥
(1)
𝑑𝑦 𝑑𝑥 =
( 𝑥 2 )
Simplify terms
𝑑𝑦 𝑑𝑥 = 1
− 𝑙𝑛5𝑥
𝑥 2 8E

26. Teachings for Exercise 8F

27. Showing how differentiation works (sort of!)
y = x2
Showing how differentiation works (sort of!)
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= (𝑥+𝛿𝑥 ) 2 − 𝑥 2 𝑥+𝛿𝑥−𝑥
Multiply the bracket
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑥 2 +2(𝛿𝑥)(𝑥)+(𝛿𝑥 ) 2 − 𝑥 2 𝑥+𝛿𝑥−𝑥
This is a lowercase ‘delta’, representing a small increase in x
Gradient = change in y ÷ change in x
Group some terms
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 2(𝛿𝑥)(𝑥)+(𝛿𝑥 ) 2 𝛿𝑥
Cancel δx
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡=2𝑥+𝛿𝑥
(x+δx, (x+δx)2)
(x+δx)2 – x2
At the original point, δx = 0
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡=2𝑥
x+δx - x
(x, x2) 8F

28. Showing how differentiation works (sort of!)
y = f(x)
Showing how differentiation works (sort of!)
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑓(𝑥+𝛿𝑥)−𝑓(𝑥) 𝑥+𝛿𝑥−𝑥
Simplify the denominator
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑓 𝑥+𝛿𝑥 −𝑓(𝑥) 𝛿𝑥
This is a lowercase ‘delta’, representing a small increase in x
Gradient = change in y ÷ change in x
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑓 𝑥+𝛿𝑥 −𝑓(𝑥) 𝛿𝑥
(x+δx, f(x+δx))
f(x+δx) – f(x)
x+δx - x
This is the ‘definition’ for differentiating a function
(x, f(x)) 8F

29. You need to be able to differentiate Trigonometric Functions
Differentiation
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑓 𝑥+𝛿𝑥 −𝑓(𝑥) 𝛿𝑥
You need to be able to differentiate Trigonometric Functions
Let f(x) = sinx A
(Angle x is in radians)
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑠𝑖𝑛 𝑥+𝛿𝑥 −𝑠𝑖𝑛(𝑥) 𝛿𝑥 r
Multiply the function using sin(A + B) x B r
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝛿𝑥+𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝛿𝑥−𝑠𝑖𝑛𝑥 𝛿𝑥 O
As δx  0
Cosδx  1
Sinδx  δx
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑠𝑖𝑛𝑥(1)+𝑐𝑜𝑠𝑥(𝛿𝑥)−𝑠𝑖𝑛𝑥 𝛿𝑥
1 2 𝑟 2 𝑥
Area of the sector OAB:
Simplify terms
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥𝛿𝑥−𝑠𝑖𝑛𝑥 𝛿𝑥
1 2 𝑟 2 𝑠𝑖𝑛𝑥
Area of the triangle OAB:
Sinx’s cancel out
As x approaches 0, the area of the triangle and sector become equal. Hence:
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑐𝑜𝑠𝑥𝛿𝑥 𝛿𝑥
1 2 𝑟 2 𝑠𝑖𝑛𝑥= 1 2 𝑟 2 𝑥
Cancel δx’s
𝑓 ′ 𝑥 =𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥=𝑥 8F

30. You need to be able to differentiate Trigonometric Functions
Differentiation
You need to be able to differentiate Trigonometric Functions
Gradient = -1
y = Cosθ
If:
𝑦=𝑠𝑖𝑛𝑥 1
𝑑𝑦 𝑑𝑥 =𝑐𝑜𝑠𝑥
y = Sinθ
Then:
π/2 π
3π/2 2π -1
If:
𝑦=𝑠𝑖𝑛𝑓 𝑥
Gradient = 0
𝑑𝑦 𝑑𝑥 =𝑓′(𝑥)𝑐𝑜𝑠𝑓(𝑥)
Then:
This means that the Cos graph is actually telling you the gradient of the Sin graph at the equivalent point!
At π, Cosθ = -1
The gradient of Sinθ at π is -1
At 3π/2, Cosθ = 0
The gradient of Sinθ at 3π/2 is 0! 8F

31. Differentiate using the chain rule
𝑦=𝑠𝑖𝑛𝑥
If:
𝑦=𝑠𝑖𝑛𝑓 𝑥
Differentiation
𝑑𝑦 𝑑𝑥 =𝑐𝑜𝑠𝑥
𝑑𝑦 𝑑𝑥 =𝑓′(𝑥)𝑐𝑜𝑠𝑓(𝑥)
Then:
Then:
Differentiate: a)
𝑦=𝑠𝑖𝑛3𝑥 c)
𝑦=𝑠𝑖 𝑛 2 𝑥
Differentiate
Rewrite
𝑑𝑦 𝑑𝑥 =3𝑐𝑜𝑠3𝑥
𝑦=(𝑠𝑖𝑛𝑥 ) 2
Differentiate using the chain rule
𝑑𝑦 𝑑𝑥 =2(𝑠𝑖𝑛𝑥 ) 1 (𝑐𝑜𝑠𝑥)
𝑦=𝑠𝑖𝑛 2 3 𝑥 b)
Simplify
Differentiate
𝑑𝑦 𝑑𝑥 =2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
𝑑𝑦 𝑑𝑥 = 2 3 𝑐𝑜𝑠 2 3 𝑥 8F

32. Teachings for Exercise 8G

33. You need to be able to differentiate Trigonometric Functions
Differentiation
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑓 𝑥+𝛿𝑥 −𝑓(𝑥) 𝛿𝑥
You need to be able to differentiate Trigonometric Functions
Let f(x) = cosx
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑐𝑜𝑠 𝑥+𝛿𝑥 −𝑐𝑜𝑠(𝑥) 𝛿𝑥
Multiply the function using cos(A + B)
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝛿𝑥−𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝛿𝑥−𝑐𝑜𝑠𝑥 𝛿𝑥
As δx  0
Cosδx  1
Sinδx  δx
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑐𝑜𝑠𝑥 1 −𝑠𝑖𝑛𝑥(𝛿𝑥)−𝑐𝑜𝑠𝑥 𝛿𝑥
Simplify terms
𝑓′(𝑥)= lim 𝛿𝑥→0 𝑐𝑜𝑠𝑥−𝑠𝑖𝑛𝑥𝛿𝑥−𝑐𝑜𝑠𝑥 𝛿𝑥
Cosx’s cancel out
𝑓′(𝑥)= lim 𝛿𝑥→0 −𝑠𝑖𝑛𝑥𝛿𝑥 𝛿𝑥
Cancel δx’s
𝑓 ′ 𝑥 =−𝑠𝑖𝑛𝑥 8G

34. You need to be able to differentiate Trigonometric Functions
Differentiation
You need to be able to differentiate Trigonometric Functions
If:
𝑦=𝑐𝑜𝑠𝑥
𝑑𝑦 𝑑𝑥 =−𝑠𝑖𝑛𝑥
Then:
If:
𝑦=𝑐𝑜𝑠⁡𝑓 𝑥
𝑑𝑦 𝑑𝑥 =− 𝑓 ′ 𝑥 𝑠𝑖𝑛𝑓(𝑥)
Then: 8G

35. Differentiate using the chain rule
𝑦=𝑐𝑜𝑠𝑥
If:
𝑦=𝑐𝑜𝑠⁡𝑓 𝑥
Differentiation
𝑑𝑦 𝑑𝑥 =−𝑠𝑖𝑛𝑥
𝑑𝑦 𝑑𝑥 =− 𝑓 ′ 𝑥 𝑠𝑖𝑛𝑓(𝑥)
Then:
Then:
Differentiate: a)
𝑦=𝑐𝑜𝑠⁡(4𝑥−3) c)
𝑦=3𝑐𝑜𝑠𝑥
Differentiate
Differentiate
𝑑𝑦 𝑑𝑥 =−4𝑠𝑖𝑛⁡(4𝑥−3)
𝑑𝑦 𝑑𝑥 =3(−𝑠𝑖𝑛𝑥)
Simplify
𝑑𝑦 𝑑𝑥 =−3𝑠𝑖𝑛𝑥 b)
𝑦=𝑐𝑜 𝑠 3 𝑥
Rewrite
𝑦=(𝑐𝑜𝑠𝑥 ) 3
Differentiate using the chain rule
𝑑𝑦 𝑑𝑥 =3(𝑐𝑜𝑠𝑥 ) 2
(−𝑠𝑖𝑛𝑥)
Rewrite/simplify
𝑑𝑦 𝑑𝑥 =−3𝑐𝑜 𝑠 2 𝑥𝑠𝑖𝑛𝑥 8G

36. Teachings for Exercise 8H

37. You need to be able to differentiate Trigonometric Functions
Differentiation
You need to be able to differentiate Trigonometric Functions
𝑢=𝑠𝑖𝑛𝑥
𝑣=𝑐𝑜𝑠𝑥
Differentiate
Differentiate
𝑦= 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥
𝑑𝑢 𝑑𝑥 =𝑐𝑜𝑠𝑥
𝑑𝑣 𝑑𝑥 =−𝑠𝑖𝑛𝑥
Let:
𝑦=𝑡𝑎𝑛𝑥
So:
𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2
Substitute values
𝑑𝑦 𝑑𝑥 =
(𝑐𝑜𝑠𝑥)
(𝑐𝑜𝑠𝑥)
− (𝑠𝑖𝑛𝑥)
(−𝑠𝑖𝑛𝑥)
(𝑐𝑜𝑠𝑥 ) 2
Simplify
𝑑𝑦 𝑑𝑥 =
𝑐𝑜 𝑠 2 𝑥
+ 𝑠𝑖 𝑛 2 𝑥
𝑐𝑜 𝑠 2 𝑥
Replace the numerator using another identity
𝑑𝑦 𝑑𝑥 = 1
𝑐𝑜 𝑠 2 𝑥
Rewrite
𝑑𝑦 𝑑𝑥 =
𝑠𝑒 𝑐 2 𝑥 8H

38. You need to be able to differentiate Trigonometric Functions
Differentiation
You need to be able to differentiate Trigonometric Functions
If:
𝑦=𝑡𝑎𝑛𝑥
𝑑𝑦 𝑑𝑥 =𝑠𝑒 𝑐 2 𝑥
Then:
If:
𝑦=𝑡𝑎𝑛⁡𝑓 𝑥
𝑑𝑦 𝑑𝑥 = 𝑓 ′ 𝑥 𝑠𝑒 𝑐 2 𝑓(𝑥)
Then: 8H

39. Differentiate using the Chain rule
𝑦=𝑡𝑎𝑛𝑥
If:
𝑦=𝑡𝑎𝑛⁡𝑓 𝑥
Differentiation
𝑑𝑦 𝑑𝑥 =𝑠𝑒 𝑐 2 𝑥
𝑑𝑦 𝑑𝑥 = 𝑓 ′ 𝑥 𝑠𝑒 𝑐 2 𝑓(𝑥)
Then:
Then:
Differentiate:
𝑦=𝑡𝑎 𝑛 4 𝑥
𝑦=(𝑡𝑎𝑛𝑥 ) 4
Differentiate using the Chain rule
𝑑𝑦 𝑑𝑥 =4(𝑡𝑎𝑛𝑥 ) 3
(𝑠𝑒 𝑐 2 𝑥)
Simplify
𝑑𝑦 𝑑𝑥 =4𝑡𝑎 𝑛 3 𝑥𝑠𝑒 𝑐 2 𝑥 8H

40. Differentiation 8H 𝑢=𝑥 𝑣=𝑡𝑎𝑛2𝑥 𝑦=𝑥𝑡𝑎𝑛2𝑥 𝑑𝑢 𝑑𝑥 =1 𝑑𝑣 𝑑𝑥 =2𝑠𝑒 𝑐 2 2𝑥
𝑦=𝑡𝑎𝑛𝑥
If:
𝑦=𝑡𝑎𝑛⁡𝑓 𝑥
Differentiation
𝑑𝑦 𝑑𝑥 =𝑠𝑒 𝑐 2 𝑥
𝑑𝑦 𝑑𝑥 = 𝑓 ′ 𝑥 𝑠𝑒 𝑐 2 𝑓(𝑥)
Then:
Then:
Differentiate:
𝑢=𝑥
𝑣=𝑡𝑎𝑛2𝑥
𝑦=𝑥𝑡𝑎𝑛2𝑥
Differentiate
Differentiate
Use the product rule
𝑑𝑢 𝑑𝑥 =1
𝑑𝑣 𝑑𝑥 =2𝑠𝑒 𝑐 2 2𝑥
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
Sub in values
𝑑𝑦 𝑑𝑥 =
(𝑥)
(2𝑠𝑒 𝑐 2 2𝑥)
+ (𝑡𝑎𝑛2𝑥)
(1)
Simplify
𝑑𝑦 𝑑𝑥 =
2𝑥𝑠𝑒 𝑐 2 2𝑥
+ 𝑡𝑎𝑛2𝑥 8H

41. Teachings for Exercise 8I

42. Differentiation 8I 𝑦=𝑐𝑜𝑠𝑒𝑐𝑥 If: 𝑦=𝑐𝑜𝑠𝑒𝑐𝑥 𝑦= 1 𝑠𝑖𝑛𝑥 𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥
You need to know how to differentiate the last 3 Trigonometric functions
𝑦=𝑐𝑜𝑠𝑒𝑐𝑥
Rewrite
If:
𝑦=𝑐𝑜𝑠𝑒𝑐𝑥
𝑦= 1 𝑠𝑖𝑛𝑥
𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥
Rewrite
Then:
𝑦=(𝑠𝑖𝑛𝑥 ) −1
Differentiate using the Chain Rule
𝑑𝑦 𝑑𝑥 =−(𝑠𝑖𝑛𝑥 ) −2
(𝑐𝑜𝑠𝑥)
If:
𝑦=𝑐𝑜𝑠𝑒𝑐⁡𝑓 𝑥
Rewrite
𝑑𝑦 𝑑𝑥 = −𝑓 ′ 𝑥 𝑐𝑜𝑠𝑒𝑐𝑓(𝑥)𝑐𝑜𝑡𝑓(𝑥)
𝑑𝑦 𝑑𝑥 =
− 𝑐𝑜𝑠𝑥 𝑠𝑖 𝑛 2 𝑥
Then:
Imagine as 2 separate fractions
𝑑𝑦 𝑑𝑥 =
− 1 𝑠𝑖𝑛𝑥
× 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥
Rewrite them
𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥 8I

43. Differentiation 8I 𝑦=𝑠𝑒𝑐𝑥 If: 𝑦=𝑠𝑒𝑐𝑥 𝑦= 1 𝑐𝑜𝑠𝑥 𝑑𝑦 𝑑𝑥 =𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 Then:
You need to know how to differentiate the last 3 Trigonometric functions
𝑦=𝑠𝑒𝑐𝑥
Rewrite
If:
𝑦=𝑠𝑒𝑐𝑥
𝑦= 1 𝑐𝑜𝑠𝑥
𝑑𝑦 𝑑𝑥 =𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
Rewrite
Then:
𝑦=(𝑐𝑜𝑠𝑥 ) −1
Differentiate using the Chain Rule
𝑑𝑦 𝑑𝑥 =−(𝑐𝑜𝑠𝑥 ) −2
(−𝑠𝑖𝑛𝑥)
If:
𝑦=𝑠𝑒𝑐⁡𝑓 𝑥
Rewrite
𝑑𝑦 𝑑𝑥 = 𝑓 ′ 𝑥 𝑠𝑒𝑐𝑓 𝑥 𝑡𝑎𝑛𝑓(𝑥)
𝑑𝑦 𝑑𝑥 =
𝑠𝑖𝑛𝑥 𝑐𝑜 𝑠 2 𝑥
Then:
Imagine as 2 separate fractions
𝑑𝑦 𝑑𝑥 =
1 𝑐𝑜𝑠𝑥
× 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥
Rewrite them
𝑑𝑦 𝑑𝑥 =𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 8I

44. Differentiation 8I 𝑦=𝑐𝑜𝑡𝑥 If: 𝑦=𝑐𝑜𝑡𝑥 𝑦= 1 𝑡𝑎𝑛𝑥 𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒 𝑐 2 𝑥
You need to know how to differentiate the last 3 Trigonometric functions
𝑦=𝑐𝑜𝑡𝑥
Rewrite
If:
𝑦=𝑐𝑜𝑡𝑥
𝑦= 1 𝑡𝑎𝑛𝑥
𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒 𝑐 2 𝑥
Rewrite
Then:
𝑦=(𝑡𝑎𝑛𝑥 ) −1
Differentiate using the Chain Rule
𝑑𝑦 𝑑𝑥 =−(𝑡𝑎𝑛𝑥 ) −2
(𝑠𝑒 𝑐 2 𝑥)
If:
𝑦=𝑐𝑜𝑡⁡𝑓 𝑥
Rewrite
𝑑𝑦 𝑑𝑥 = −𝑓 ′ 𝑥 𝑐𝑜𝑠𝑒 𝑐 2 𝑓(𝑥)
𝑑𝑦 𝑑𝑥 =
− 𝑠𝑒 𝑐 2 𝑥 𝑡𝑎 𝑛 2 𝑥
Then:
Imagine as 2 separate fractions
𝑑𝑦 𝑑𝑥 =
− 𝑠𝑒 𝑐 2 𝑥 1
× 1 𝑡𝑎 𝑛 2 𝑥
𝑑𝑦 𝑑𝑥 =− 1 𝑠𝑖 𝑛 2 𝑥
Rewrite them
𝑑𝑦 𝑑𝑥 =
− 1 𝑐𝑜 𝑠 2 𝑥
× 𝑐𝑜 𝑠 2 𝑥 𝑠𝑖 𝑛 2 𝑥
Rewrite
𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒 𝑐 2 𝑥 8I

45. Differentiate using the Chain rule
𝑦=𝑐𝑜𝑠𝑒𝑐𝑥
If:
𝑦=𝑠𝑒𝑐𝑥
If:
𝑦=𝑐𝑜𝑡𝑥
𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥
𝑑𝑦 𝑑𝑥 =𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒 𝑐 2 𝑥
Then:
Then:
Then:
If:
𝑦=𝑐𝑜𝑠𝑒𝑐⁡𝑓 𝑥
If:
𝑦=𝑠𝑒𝑐⁡𝑓 𝑥
If:
𝑦=𝑐𝑜𝑡⁡𝑓 𝑥
𝑑𝑦 𝑑𝑥 = −𝑓 ′ 𝑥 𝑐𝑜𝑠𝑒𝑐𝑓(𝑥)𝑐𝑜𝑡𝑓(𝑥)
𝑑𝑦 𝑑𝑥 = 𝑓 ′ 𝑥 𝑠𝑒𝑐𝑓 𝑥 𝑡𝑎𝑛𝑓(𝑥)
𝑑𝑦 𝑑𝑥 = −𝑓 ′ 𝑥 𝑐𝑜𝑠𝑒 𝑐 2 𝑓(𝑥)
Then:
Then:
Then:
Differentiate:
𝑦=𝑠𝑒 𝑐 3 𝑥
Rewrite
𝑦=(𝑠𝑒𝑐𝑥 ) 3
Differentiate using the Chain rule
𝑑𝑦 𝑑𝑥 =3(𝑠𝑒𝑐𝑥 ) 2
(𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥)
Simplify
𝑑𝑦 𝑑𝑥 =3𝑠𝑒 𝑐 3 𝑥𝑡𝑎𝑛𝑥 8I

46. 8I Differentiate: 𝑦= 𝑐𝑜𝑠𝑒𝑐2𝑥 𝑥 2 𝑢=𝑐𝑜𝑠𝑒𝑐2𝑥 𝑣= 𝑥 2 𝑑𝑣 𝑑𝑥 =2𝑥
𝑦=𝑐𝑜𝑠𝑒𝑐𝑥
If:
𝑦=𝑠𝑒𝑐𝑥
If:
𝑦=𝑐𝑜𝑡𝑥
𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥
𝑑𝑦 𝑑𝑥 =𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒 𝑐 2 𝑥
Then:
Then:
Then:
If:
𝑦=𝑐𝑜𝑠𝑒𝑐⁡𝑓 𝑥
If:
𝑦=𝑠𝑒𝑐⁡𝑓 𝑥
If:
𝑦=𝑐𝑜𝑡⁡𝑓 𝑥
𝑑𝑦 𝑑𝑥 = −𝑓 ′ 𝑥 𝑐𝑜𝑠𝑒𝑐𝑓(𝑥)𝑐𝑜𝑡𝑓(𝑥)
𝑑𝑦 𝑑𝑥 = 𝑓 ′ 𝑥 𝑠𝑒𝑐𝑓 𝑥 𝑡𝑎𝑛𝑓(𝑥)
𝑑𝑦 𝑑𝑥 = −𝑓 ′ 𝑥 𝑐𝑜𝑠𝑒 𝑐 2 𝑓(𝑥)
Then:
Then:
Then:
Differentiate:
𝑦= 𝑐𝑜𝑠𝑒𝑐2𝑥 𝑥 2
𝑢=𝑐𝑜𝑠𝑒𝑐2𝑥
𝑣= 𝑥 2
Use the quotient rule
Differentiate
Differentiate
𝑑𝑣 𝑑𝑥 =2𝑥
𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2
𝑑𝑢 𝑑𝑥 =−2𝑐𝑜𝑠𝑒𝑐2𝑥𝑐𝑜𝑡2𝑥
𝑑𝑦 𝑑𝑥 =
( 𝑥 2 )
(−2𝑐𝑜𝑠𝑒𝑐2𝑥𝑐𝑜𝑡2𝑥)
− (𝑐𝑜𝑠𝑒𝑐2𝑥)
(2𝑥)
( 𝑥 2 ) 2
Simplify brackets
𝑑𝑦 𝑑𝑥 =
−2 𝑥 2 𝑐𝑜𝑠𝑒𝑐2𝑥𝑐𝑜𝑡2𝑥
− 2𝑥𝑐𝑜𝑠𝑒𝑐2𝑥
𝑥 4
Factorise the numerator
𝑑𝑦 𝑑𝑥 =
−2𝑥𝑐𝑜𝑠𝑒𝑐2𝑥
(𝑥𝑐𝑜𝑡2𝑥 +1)
𝑥 4
Remove common factors to the numerator and denominator
𝑑𝑦 𝑑𝑥 =
−2𝑐𝑜𝑠𝑒𝑐2𝑥
(𝑥𝑐𝑜𝑡2𝑥 +1)
𝑥 3 8I

47. If:
𝑦=𝑐𝑜𝑠𝑒𝑐𝑥
If:
𝑦=𝑠𝑒𝑐𝑥
If:
𝑦=𝑐𝑜𝑡𝑥
𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥
𝑑𝑦 𝑑𝑥 =𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒 𝑐 2 𝑥
Then:
Then:
Then:
If:
𝑦=𝑐𝑜𝑠𝑒𝑐⁡𝑓 𝑥
If:
𝑦=𝑠𝑒𝑐⁡𝑓 𝑥
If:
𝑦=𝑐𝑜𝑡⁡𝑓 𝑥
𝑑𝑦 𝑑𝑥 = −𝑓 ′ 𝑥 𝑐𝑜𝑠𝑒𝑐𝑓(𝑥)𝑐𝑜𝑡𝑓(𝑥)
𝑑𝑦 𝑑𝑥 = 𝑓 ′ 𝑥 𝑠𝑒𝑐𝑓 𝑥 𝑡𝑎𝑛𝑓(𝑥)
𝑑𝑦 𝑑𝑥 = −𝑓 ′ 𝑥 𝑐𝑜𝑠𝑒 𝑐 2 𝑓(𝑥)
Then:
Then:
Then:
This is all the differentiation you get in the formula booklet, the rest you must learn! 8I

48. Teachings for Exercise 8J

49. Differentiation 8J 𝑢= 𝑒 𝑥 𝑣=𝑠𝑖𝑛𝑥 𝑦= 𝑒 𝑥 𝑠𝑖𝑛𝑥 𝑑𝑢 𝑑𝑥 = 𝑒 𝑥 𝑑𝑣 𝑑𝑥 =𝑐𝑜𝑠𝑥
Differentiate:
𝑢= 𝑒 𝑥
𝑣=𝑠𝑖𝑛𝑥
𝑦= 𝑒 𝑥 𝑠𝑖𝑛𝑥
Differentiate
Differentiate
Use the product rule
𝑑𝑢 𝑑𝑥 = 𝑒 𝑥
𝑑𝑣 𝑑𝑥 =𝑐𝑜𝑠𝑥
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
Sub in values
𝑑𝑦 𝑑𝑥 =
(𝑒 𝑥 )
(𝑐𝑜𝑠𝑥)
+ (𝑠𝑖𝑛𝑥)
( 𝑒 𝑥 )
Simplify
𝑑𝑦 𝑑𝑥 =
𝑒 𝑥 𝑐𝑜𝑠𝑥
+ 𝑒 𝑥 𝑠𝑖𝑛𝑥
Factorise
𝑑𝑦 𝑑𝑥 =
𝑒 𝑥 (𝑐𝑜𝑠𝑥+𝑠𝑖𝑛𝑥) 8J

50. Rewrite lnxcosx so it can be grouped (multiply top and bottom by x)
Differentiation
Differentiate:
𝑢=𝑙𝑛𝑥
𝑣=𝑠𝑖𝑛𝑥
𝑦= 𝑙𝑛𝑥 𝑠𝑖𝑛𝑥
Differentiate
Differentiate
𝑑𝑢 𝑑𝑥 = 1 𝑥
𝑑𝑣 𝑑𝑥 =𝑐𝑜𝑠𝑥
Use the quotient rule
𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2
1 𝑥
Sub in values
(𝑠𝑖𝑛𝑥)
− (𝑙𝑛𝑥)
(𝑐𝑜𝑠𝑥)
𝑑𝑦 𝑑𝑥 =
(𝑠𝑖𝑛𝑥 ) 2
Simplify
𝑠𝑖𝑛𝑥 𝑥
− 𝑙𝑛𝑥𝑐𝑜𝑠𝑥
𝑑𝑦 𝑑𝑥 =
𝑠𝑖 𝑛 2 𝑥
Rewrite lnxcosx so it can be grouped (multiply top and bottom by x)
𝑠𝑖𝑛𝑥−𝑥𝑙𝑛𝑥𝑐𝑜𝑠𝑥 𝑥
𝑑𝑦 𝑑𝑥 =
𝑠𝑖 𝑛 2 𝑥
Simplify
𝑑𝑦 𝑑𝑥 =
𝑠𝑖𝑛𝑥−𝑥𝑙𝑛𝑥𝑐𝑜𝑠𝑥 𝑥𝑠𝑖 𝑛 2 𝑥 8J

51. Summary
We have learnt how to differentiate combined functions using the Chain, Product and Quotient Rules
We have also seen how to differentiate the exponential, logarithmic and trigonometric functions
Remember that all this can be used alongside finding equations of tangents, normals and turning points!