Differentiation - Topic 7

Question 1

Year 1 - Chapter 12.2

What is the equation for differentiating by first principles?

Answer 1

f′(x) =

limh→0 (f(x+h)−f(x)h) ÷ h,

h ≠ 0

Question 2

Year 1 - Chapter 12.3

Differentiate:
1. y = xⁿ
2. f(x) = xⁿ
3. y = axⁿ
4. f(x) = axⁿ

Answer 2

1. dy/dx = nxⁿ⁻¹
2. f’(x) = nxⁿ⁻¹
3. dy/dx = anxⁿ⁻¹
4. f’(x) = anxⁿ⁻¹

Question 3

Year 1 - Chapter 12.4

How to differentiate a quadratic step-by-step

Equation: y = ax² + bx + c

Answer 3

1. ax² → 2ax¹ ≡ 2ax
   The quadratic tells you the slope of the gradient function
2. bx ≡ bx¹ → 1bx⁰ ≡ bx
   An x term differeniates to give a constant
3. c → 0
   Constant terms disappear when you differentiate

Question 4

Year 1 - Chapter 12.6

List of equations for the tangent and normal of the curve y = f(x)

Answer 4

Tangent:
y - f(a) = f’(a)(x-a)

Normal:
y - f(a) = -(1÷f’(a))(x-a)

Question 5

Year 1 - Chapter 12.6

Differentiate the equation of the tangent to the curve

Equation: y = x³ - 3x² + 2x - 1, at the point (3,5)

Answer 5

y = x³ - 3x² + 2x - 1
↓
dy/dx = 3x² - 6x + 2
↓
When x = 3;
3(3)² - 6(3) + 2 = 11
↓
y₁ - y₂ = m(x₁ - x₂)
y - 5 = 11(x - 3)
↓
y = 11x - 28

Question 6

Year 1 - Chapter 12.7

How can you use the derivative to determine whether a function is increasing or decreasing?

Answer 6

• The function f(x) is increasing on the interval [a,b] if f’(x) ≥ 0 for all values such that a<x<b
• The function f(x) is increasing on the interval [a,b] if f’(x) ≤ 0 for all values such that a<x<b

Question 7

Year 1 - Chapter 12.8

Differentiate by the second order

Answer 7

y = 5x³
↓
dy/dx = 15x²
↓
d²y/dx² = 30x

d²y/dx² is the same as f’‘(x)

Question 8

Year 1 - Chapter 12.9

How do you determine the nature of a stationary point?

Answer 8

By using the second derivative;
* If f”(a) < 0; the point is a local maximum
* If f”(a) > 0; the point is a local minimum
* If f”(a) = 0; the point could be either, you must look at points next to it to discern whether its the max or min

Question 9

Year 1 - Chapter 12.10

How to sketch from y = f(x) → y = f’(x)

Answer 9

• Maximum or minimum points → cut the x-axis
• Point of inflection → touches the x-axis
• Positive gradient → above the x-axis
• Negative gradient → below the x-axis
• Vertical asymptote → vertical asymptote
• Horizontal asymptote → horizontal asymptote at the x-axis

Question 10

Year 1 - Chapter 12.11

How can differentiate be used in modelling questions?

Answer 10

dy/dx ≡ small change in y/small change in x

It represents a rate of change of y in respect to x

If V represented volume of water, and t the time, you can model V as a function of t.
* V = f(t) → dV/dt = f’(t), which represents the change in volume

Question 11

Year 2 - Chapter 9.1

How to differentiate sin(x) and cos(x) by small angle approximations in radians

Answer 11

sin(x) ≈ x
cos(x) ≈ 1 - 1/2(x²)

Question 12

Year 2 - Chapter 9.1

How to differentiate sin(x) and cos(x) by first principles in radians

Answer 12

• limh→0 sinh/h = 1
• limh→0 (cosh - 1)/h = 0

In-depth proof for cos(x) by first principles in book, pg. 232

Question 13

Year 2 - Chapter 9.1

Differentiate:
1. y = sin(kx)
2. y = cos(kx)

Answer 13

1. dy/dx = kcos(kx)
2. dy.dx = -ksin(kx)

Question 14

Year 9 - Chapter 9.2

List of differentials of exponentials and logarithms:

Answer 14

• If y = eᵏˣ → dy/dx = keᵏˣ
• If y = ln(x) → dy/dx = 1/x
• If y = aᵏˣ (where k is a real constant and a > 0) → dy/dx = aᵏˣln(a)

Question 15

Year 2 - Chapter 9.3

Equation for the chain rule?

Answer 15

dy/dx = dy/du x du/dx

Question 16

Year 2 - Chapter 9.3

How to use the chain rule

Example

Answer 16

y = (3x⁴ + x)⁵

u = 3x⁴ + x
du/dx = 12x³ + 1
↓
y = u⁵
dy/du = 5u⁴
↓
dy/dx = dy/du x du/dx
dy/dx = 5u⁴(12x³ + 1)
↓
dy/dx = 5(3x⁴ + x)⁴(12x³ + 1)

Question 17

Year 2 - Chapter 9.3

How do you get the first derivative if the equation is e.g., y³ + y = x

At the point (2,1)

Answer 17

dy/dx = 1 ÷ dx/dy

dx/dy = 3y² + 1
↓
dy/dx = 1/(3y² + 1)
↓
Substitute (2,1) in;
1/(3(1)² + 1) = 1/4

Question 18

Year 2 - Chapter 9.4

Equation for the product rule?

Answer 18

If y = uv (when both u and v are functions of x)

dy/dx = u(dv/dx) + v(du/dx)

Question 19

Year 2 - Chapter 9.4

How to use the product rule

Example

Answer 19

f(x) = x²√(3x-1)

u = x²
v = √(3x-1) → (3x-1)¹/²
↓
du/dx = 2x
dv/dx = 3/2(3x-1)⁻¹/²
↓
dy/dx = u(dv/dx) + v(du/dx)
dy/dx = x²(3/2(3x-1)⁻¹/²) + √(3x-1)(2x)
↓
dy/dx = x(15x-4) ÷ 2√(3x-1)

Use the chain rule for (3x-1)¹/²

Question 20

Year 2 - Chapter 9.5

Equation for the quotient rule?

Answer 20

If y = u/v (when both u and v are functions of x)

dy/dx = (v(du/dx) - (u(dv/dx)) ÷ v²

Question 21

Year 2 - Chapter 9.5

How to use the quotient rule

Example

Answer 21

y = x/(2x+5)

u = x
v = 2x+5
↓
du/dx = 1
dv/dx = 2
↓
dy/dx = (v(du/dx) - (u(dv/dx)) ÷ v²
dy/dx = (2x+5(1) - (x(2) ÷ (2x+5)²
↓
dy/dx = 5/2(x+5)²