Differentiation - Topic 7 Flashcards
f(x), x^n, applying differentiation, differential rules, simple functions, and simple differential equations in context
Year 1 - Chapter 12.2
What is the equation for differentiating by first principles?
f′(x) =
limh→0 (f(x+h)−f(x)h) ÷ h,
h ≠ 0
Year 1 - Chapter 12.3
Differentiate:
1. y = xⁿ
2. f(x) = xⁿ
3. y = axⁿ
4. f(x) = axⁿ
- dy/dx = nxⁿ⁻¹
- f’(x) = nxⁿ⁻¹
- dy/dx = anxⁿ⁻¹
- f’(x) = anxⁿ⁻¹
Year 1 - Chapter 12.4
How to differentiate a quadratic step-by-step
Equation: y = ax² + bx + c
- ax² → 2ax¹ ≡ 2ax
The quadratic tells you the slope of the gradient function - bx ≡ bx¹ → 1bx⁰ ≡ bx
An x term differeniates to give a constant - c → 0
Constant terms disappear when you differentiate
Year 1 - Chapter 12.6
List of equations for the tangent and normal of the curve y = f(x)
Tangent:
y - f(a) = f’(a)(x-a)
Normal:
y - f(a) = -(1÷f’(a))(x-a)
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)
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
Year 1 - Chapter 12.7
How can you use the derivative to determine whether a function is increasing or decreasing?
- 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
Year 1 - Chapter 12.8
Differentiate by the second order
y = 5x³
↓
dy/dx = 15x²
↓
d²y/dx² = 30x
d²y/dx² is the same as f’‘(x)
Year 1 - Chapter 12.9
How do you determine the nature of a stationary point?
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
Year 1 - Chapter 12.10
How to sketch from y = f(x) → y = f’(x)
- 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
Year 1 - Chapter 12.11
How can differentiate be used in modelling questions?
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
Year 2 - Chapter 9.1
How to differentiate sin(x) and cos(x) by small angle approximations in radians
sin(x) ≈ x
cos(x) ≈ 1 - 1/2(x²)
Year 2 - Chapter 9.1
How to differentiate sin(x) and cos(x) by first principles in radians
- limh→0 sinh/h = 1
- limh→0 (cosh - 1)/h = 0
In-depth proof for cos(x) by first principles in book, pg. 232
Year 2 - Chapter 9.1
Differentiate:
1. y = sin(kx)
2. y = cos(kx)
- dy/dx = kcos(kx)
- dy.dx = -ksin(kx)
Year 9 - Chapter 9.2
List of differentials of exponentials and logarithms:
- 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)
Year 2 - Chapter 9.3
Equation for the chain rule?
dy/dx = dy/du x du/dx
Year 2 - Chapter 9.3
How to use the chain rule
Example
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)
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)
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
Year 2 - Chapter 9.4
Equation for the product rule?
If y = uv (when both u and v are functions of x)
dy/dx = u(dv/dx) + v(du/dx)
Year 2 - Chapter 9.4
How to use the product rule
Example
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
Year 2 - Chapter 9.5
Equation for the quotient rule?
If y = u/v (when both u and v are functions of x)
dy/dx = (v(du/dx) - (u(dv/dx)) ÷ v²
Year 2 - Chapter 9.5
How to use the quotient rule
Example
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