1. Differentiation

2. After completing this chapter you should be able to:
Find the gradient of a curve whose equation is expressed in a parametric form
Differentiate implicit relationships
Differentiate power functions such as ax
Use the chain rule to connect the rates of change of two variables
Set up simple differential equations from information given in context

3. 4.1 You can find the gradient of a curve given in parametric coordinates
You differentiate x and y with respect to t
Then you use the chain rule rearranged into the form
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 ÷ 𝑑𝑥 𝑑𝑡

4. 1. Find the gradient of the curve with parametric equations
x = t², y = 2t³
𝑑𝑥 𝑑𝑡 =2𝑡
𝑑𝑦 𝑑𝑡 =6 𝑡 2
𝑑𝑦 𝑑𝑥 =6 𝑡 2 ÷2𝑡=3𝑡
2. Find the gradient of the curve with parametric equations
x = 𝑒 𝑡 + 𝑒 −𝑡 , y = 𝑒 𝑡 − 𝑒 −𝑡
𝑑𝑥 𝑑𝑡 = 𝑒 𝑡 − 𝑒 −𝑡
𝑑𝑦 𝑑𝑡 = 𝑒 𝑡 + 𝑒 −𝑡
𝑑𝑦 𝑑𝑥 = 𝑒 𝑡 + 𝑒 −𝑡 𝑒 𝑡 − 𝑒 −𝑡
this can be tidied up by multiplying by 𝑒 𝑡 𝑒 𝑡
𝑔𝑖𝑣𝑖𝑛𝑔 𝑒 2𝑡 +1 𝑒 2𝑡 −1

5. 3. Find the gradient of the curve with parametric equations
x =𝑠𝑖𝑛2𝑡 , y = 𝑠𝑖𝑛𝑡
𝑑𝑥 𝑑𝑡 =2 𝑐𝑜𝑠2𝑡
𝑑𝑦 𝑑𝑡 =𝑐𝑜𝑠𝑡 so
𝑑𝑦 𝑑𝑥 = 𝑐𝑜𝑠𝑡 2𝑐𝑜𝑠2𝑡
Exercise 4A page 38

6. 4.2 You can differentiate relations which are implicit, such as x² + y² = 8x, and cos(x + y) = siny
The basic skill is to differentiate each term in turn using the chain rule and the product rule as appropriate:
𝑑 𝑑𝑥 𝑦 𝑛 =𝑛 𝑦 −1 𝑑𝑦 𝑑𝑥 ①
𝑑 𝑑𝑥 𝑥𝑦 =𝑥 𝑑 𝑑𝑥 𝑦 +𝑦 𝑑 𝑑𝑥 (𝑥)
=𝑥 𝑑𝑦 𝑑𝑥 +𝑦×1
=𝑥 𝑑𝑦 𝑑𝑥 +𝑦 ②

7. Find 𝒅𝒚 𝒅𝒙 in terms of x and y where 𝒙 𝟐 +𝟑 𝒚 𝟐 −𝟔𝒙=𝟏𝟐
Differentiating term by term gives us
𝑑 𝑑𝑥 𝑥 2 =2𝑥
𝑑 𝑑𝑥 3 𝑦 2 =6𝑦 𝑑𝑦 𝑑𝑥
𝑑 𝑑𝑥 6𝑥 =6
This gives us
2𝑥+6𝑦 𝑑𝑦 𝑑𝑥 −6=0
𝑑𝑦 𝑑𝑥 = 6 −2𝑥 6𝑦
𝑑𝑦 𝑑𝑥 = 3 −𝑥 3𝑦

8. 𝑑𝑦 𝑑𝑥 = 12𝑥𝑦 −4 𝑥 3 12 𝑦 2 −6 𝑥 2 = 2𝑥 3𝑦 −2 𝑥 2 3 4 𝑦 2 −3 𝑥 2
Find 𝒅𝒚 𝒅𝒙 in terms of x and y where 𝒙 𝟒 −𝟔 𝒙 𝟐 𝒚+𝟒 𝒚 𝟑 =𝟐
Differentiating term by term
𝑑 𝑑𝑥 (𝑥 4 )=4 𝑥 3
𝑑 𝑑𝑥 6 𝑥 2 𝑦 =6 𝑥 2 𝑑𝑦 𝑑𝑥 +12𝑥𝑦
𝑑 𝑑𝑥 4 𝑦 3 =12 𝑦 2 𝑑𝑦 𝑑𝑥
From ②
From ①
This gives us
4 𝑥 3 −12𝑥𝑦 −6 𝑥 2 𝑑𝑦 𝑑𝑥 +12 𝑦 2 𝑑𝑦 𝑑𝑥 = 0
Collect all the 𝑑𝑦 𝑑𝑥 terms
12 𝑦 2 −6 𝑥 2 𝑑𝑦 𝑑𝑥 =12𝑥𝑦 −4 𝑥 3
𝑑𝑦 𝑑𝑥 = 12𝑥𝑦 −4 𝑥 𝑦 2 −6 𝑥 2 = 2𝑥 3𝑦 −2 𝑥 𝑦 2 −3 𝑥 2

9. 4.3 You can differentiate the general power function ax, where a is constant
𝑖𝑓 𝑦= 𝑎 𝑥 , 𝑡ℎ𝑒𝑛 𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 𝑥 𝑎 𝑥 𝑙𝑛𝑎
1. Find 𝑑𝑦 𝑑𝑥 𝑤ℎ𝑒𝑛 𝑦= 10 𝑥
2. Find 𝑑𝑦 𝑑𝑥 𝑤ℎ𝑒𝑛 𝑦= 5 𝑥 2
𝑑𝑦 𝑑𝑥 =2𝑥 5 𝑥 2 ln 5
𝑑𝑦 𝑑𝑥 = 10 𝑥 ln 10

10. 3. Find 𝑑𝑦 𝑑𝑥 𝑤ℎ𝑒𝑛 𝑦= 𝑥3 𝑥 𝑑𝑦 𝑑𝑥 =𝑥𝑙𝑛3. 3 𝑥 + 3 𝑥
Use the product rule
𝑑𝑦 𝑑𝑥 =𝑥𝑙𝑛3. 3 𝑥 + 3 𝑥
4. Find 𝑑𝑦 𝑑𝑥 𝑤ℎ𝑒𝑛 𝑦= 4 𝑥
𝑑𝑦 𝑑𝑥 = 𝑥 − 1 2 𝑙𝑛 𝑥
Exercise 4C page 41

11. 4.4 You can relate one rate of change to another
If a question involves two or more variables you can connect the rates of change by using the chain rule.

12. When r = 3 the value of 𝑑𝑉 𝑑𝑡 = 4𝜋 3
Given that the volume V cm³ of a hemisphere is related to its radius r cm by the formula V = 2 3 𝜋 𝑟 3 and that the hemisphere expands so that the rate of increase of it’s radius in cm-1 is given by 𝑑𝑟 𝑑𝑡 = 2 𝑟 3 , find the exact value of 𝑑𝑉 𝑑𝑡 when r = 3.
𝑑𝑉 𝑑𝑟 =2𝜋 𝑟 2
𝑑𝑟 𝑑𝑡 = 2 𝑟 3
V = 2 3 𝜋 𝑟 3
𝑑𝑉 𝑑𝑡 = 𝑑𝑉 𝑑𝑟 × 𝑑𝑟 𝑑𝑡
∴ 𝑑𝑉 𝑑𝑡 =2𝜋 𝑟 2 × 2 𝑟 3 = 4𝜋 𝑟
When r = 3 the value of 𝑑𝑉 𝑑𝑡 = 4𝜋 3

13. 4.5 You can set up a differential equation from information given in context
In Core 4 we are only going to look at first order differential equations , which only involves the first derivative

14. 𝑑𝑦 𝑑𝑥 ∝ 𝑦 2 𝑑𝑦 𝑑𝑥 =𝑘 𝑦 2 1=𝑘 𝑦 2 𝑘 = 1 9 Equation is 𝑑𝑦 𝑑𝑥 = 𝑦 2 9
A curve C has equation y = f(x) and it’s gradient at each point on the curve is directly proportional to the square of it’s y coordinate at that point. At the point (0, 3) on the curve the gradient is 1.
Write down a differential equation, which could be solved to give the equation of the curve.
State the value of any constant of proportionality which you use.
𝑑𝑦 𝑑𝑥 ∝ 𝑦 2
𝑑𝑦 𝑑𝑥 =𝑘 𝑦 2
At (0, 3) the gradient is 1
1=𝑘 𝑦 2
𝑘 = 1 9
Equation is 𝑑𝑦 𝑑𝑥 = 𝑦 2 9

15. 𝑑𝑅 𝑑𝑡 = 𝑑𝑉 𝑑𝑡 × 𝑑𝑅 𝑑𝑉 =−𝑘 4𝜋 𝑅 2 × 1 4𝜋 𝑅 2 =−𝑘
The head of a snowman of radius R cm loses volume by evaporation at a rate proportional to it’s surface area. Assuming that the head is spherical, that the volume of a sphere is 𝑉= 4 3 𝜋 𝑅 3 cm³ and that the surface is 4πR² cm², write down a differential equation for the rate of change of radius of the snowman’s head.
We need to find 𝑑𝑅 𝑑𝑡
we know 𝑑𝑉 𝑑𝑡 =−𝑘𝐴
V is the volume, A is the surface area, k is the constant of proportionality, t is time. (head is shrinking so equation is - )
and V= 4 3 𝜋 𝑅 and A = 4πR²
𝑤𝑒 𝑐𝑎𝑛 𝑓𝑖𝑛𝑑 𝑑𝑉 𝑑𝑅 =4𝜋 𝑅 2
𝑑𝑅 𝑑𝑡 = 𝑑𝑉 𝑑𝑡 × 𝑑𝑅 𝑑𝑉 =−𝑘 4𝜋 𝑅 2 × 1 4𝜋 𝑅 2 =−𝑘

16. all done
Exercise 4E page 45