Chapter 14: Calculus (Differentiation and Integration)
14.1 Introduction to Differentiation
The derivative of a function measures the rate of change of the function with respect to its variable. Geometrically, it is the gradient of the tangent to the curve at a point.
Notation:
\[
f'(x) \quad \text{or} \quad \frac{dy}{dx}
\]
Derivative of \(x^n\) (for any rational \(n\)):
\[
\frac{d}{dx}(x^n) = nx^{n-1}
\]
Derivatives of standard functions:
- \(\frac{d}{dx}(\sin x) = \cos x\)
- \(\frac{d}{dx}(\cos x) = -\sin x\)
- \(\frac{d}{dx}(\tan x) = \sec^2 x\)
- \(\frac{d}{dx}(e^x) = e^x\)
- \(\frac{d}{dx}(\ln x) = \frac{1}{x}\)
Constant multiple and sum rules:
\(\frac{d}{dx}(k f(x)) = k f'(x)\)
\(\frac{d}{dx}(f(x) + g(x)) = f'(x) + g'(x)\)
Worked Example 14.1 (Differentiation – non-calculator)
Differentiate:
- \(y = 3x^4\)
- \(y = x^5 + 2x^3 - 7x + 1\)
- \(y = \sin 2x\)
Solution:
(a) \(\frac{dy}{dx} = 12x^3\)
(b) \(\frac{dy}{dx} = 5x^4 + 6x^2 - 7\)
(c) \(\frac{dy}{dx} = 2\cos 2x\)
14.2 Product and Quotient Rules
Product rule:
\[
\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}
\]
Quotient rule:
\[
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}
\]
Worked Example 14.2 (Product and quotient rules)
Differentiate:
- \(y = x^2 e^x\)
- \(y = \frac{x}{x^2 + 1}\)
Solution:
(a) Let \(u = x^2\), \(v = e^x\)
\(\frac{du}{dx} = 2x\), \(\frac{dv}{dx} = e^x\)
\(\frac{dy}{dx} = x^2 e^x + e^x(2x) = e^x(x^2 + 2x)\)
(b) Let \(u = x\), \(v = x^2 + 1\)
\(\frac{du}{dx} = 1\), \(\frac{dv}{dx} = 2x\)
\(\frac{dy}{dx} = \frac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}\)
14.3 Chain Rule (Composite Functions)
\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)
This is used for differentiating functions of the form \(f(g(x))\).
Worked Example 14.3 (Chain rule)
Differentiate \(y = \ln(x^2 + 1)\).
Solution:
Let \(u = x^2 + 1\), then \(y = \ln u\).
\(\frac{du}{dx} = 2x\), \(\frac{dy}{du} = \frac{1}{u}\)
\(\frac{dy}{dx} = \frac{1}{u} \times 2x = \frac{2x}{x^2 + 1}\)
14.4 Applications of Differentiation
1. Gradients, Tangents and Normals
- Gradient of tangent at \(x = a\): \(m = f'(a)\)
- Equation of tangent: \(y - f(a) = f'(a)(x - a)\)
- Gradient of normal: \(m_n = -\frac{1}{f'(a)}\) (perpendicular to tangent)
Worked Example 14.4 (Tangent – non-calculator)
Find the equation of the tangent to the curve \(y = x^2 - 4x + 3\) at the point where \(x = 3\).
Solution:
When \(x = 3\): \(y = 9 - 12 + 3 = 0\). Point: \((3, 0)\).
\(\frac{dy}{dx} = 2x - 4\). At \(x = 3\), \(m = 2(3) - 4 = 2\).
Equation of tangent: \(y - 0 = 2(x - 3) \Rightarrow y = 2x - 6\)
2. Stationary Points (Turning Points)
- Solve \(f'(x) = 0\)
- First derivative test: Sign of \(f'\) around the point
- Second derivative test:
- \(f''(a) > 0\) \(\rightarrow\) minimum
- \(f''(a) < 0\) \(\rightarrow\) maximum
- \(f''(a) = 0\) \(\rightarrow\) point of inflection (or need first derivative test)
Worked Example 14.5 (Stationary points)
Find the stationary points of \(y = x^3 - 6x^2 + 9x - 1\) and determine their nature.
Solution:
\(\frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)\)
Set \(\frac{dy}{dx} = 0\): \(x = 1\) or \(x = 3\)
When \(x = 1\): \(y = 1 - 6 + 9 - 1 = 3\) \(\rightarrow\) \((1, 3)\)
When \(x = 3\): \(y = 27 - 54 + 27 - 1 = -1\) \(\rightarrow\) \((3, -1)\)
\(\frac{d^2y}{dx^2} = 6x - 12\)
At \(x = 1\): \(\frac{d^2y}{dx^2} = -6 < 0\) \(\rightarrow\) maximum at \((1, 3)\)
At \(x = 3\): \(\frac{d^2y}{dx^2} = 6 > 0\) \(\rightarrow\) minimum at \((3, -1)\)
3. Connected Rates of Change
\(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\)
Worked Example 14.6 (Mauritian context – rates of change)
The volume of a water tank is \(V = \frac{4}{3}\pi r^3\). Water is flowing into the tank at a rate of 2 m³/min. Find the rate at which the radius is increasing when \(r = 2\) m.
Solution:
\(\frac{dV}{dt} = 2\) m³/min
\(\frac{dV}{dr} = 4\pi r^2\)
\(\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} \Rightarrow 2 = 4\pi r^2 \times \frac{dr}{dt}\)
When \(r = 2\): \(2 = 4\pi(4) \times \frac{dr}{dt} = 16\pi \times \frac{dr}{dt}\)
\(\frac{dr}{dt} = \frac{2}{16\pi} = \frac{1}{8\pi}\) m/min
4. Small Increments and Approximations
\(\delta y \approx \frac{dy}{dx} \times \delta x\)
\(f(x + \delta x) \approx f(x) + f'(x)\delta x\)
5. Practical Maxima and Minima Problems
Worked Example 14.7 (Mauritian context – maximum volume)
A rectangular box with a square base has a fixed surface area of 150 cm². Find the maximum volume of the box.
Solution:
Let \(x\) = side of square base, \(h\) = height.
Surface area: \(x^2 + 4xh = 150 \Rightarrow h = \frac{150 - x^2}{4x}\)
Volume: \(V = x^2h = x^2\left(\frac{150 - x^2}{4x}\right) = \frac{150x - x^3}{4}\)
\(\frac{dV}{dx} = \frac{150 - 3x^2}{4}\)
Set \(\frac{dV}{dx} = 0\): \(150 - 3x^2 = 0 \Rightarrow x^2 = 50 \Rightarrow x = 5\sqrt{2}\)
\(\frac{d^2V}{dx^2} = -\frac{3x}{2} < 0\) when \(x > 0\), so this is a maximum.
\(V_{\text{max}} = \frac{150(5\sqrt{2}) - (5\sqrt{2})^3}{4} = \frac{750\sqrt{2} - 250\sqrt{2}}{4} = \frac{500\sqrt{2}}{4} = 125\sqrt{2} \approx 176.78\) cm³
14.5 Introduction to Integration
Integration is the reverse process of differentiation.
Indefinite integrals:
\[
\int f(x) \, dx = F(x) + C \quad \text{where } F'(x) = f(x)
\]
Power rule for integration:
\[
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)
\]
Special cases:
- \(\int \frac{1}{x} \, dx = \ln|x| + C\)
- \(\int e^x \, dx = e^x + C\)
- \(\int \sin x \, dx = -\cos x + C\)
- \(\int \cos x \, dx = \sin x + C\)
Integration of \((ax + b)^n\):
\[
\int (ax + b)^n \, dx = \frac{(ax + b)^{n+1}}{a(n+1)} + C \quad (n \neq -1)
\]
Integration of \(\sin(ax + b)\) and \(\cos(ax + b)\):
- \(\int \sin(ax + b) \, dx = -\frac{1}{a}\cos(ax + b) + C\)
- \(\int \cos(ax + b) \, dx = \frac{1}{a}\sin(ax + b) + C\)
Integration of \(e^{ax + b}\):
\[
\int e^{ax + b} \, dx = \frac{1}{a}e^{ax + b} + C
\]
Definite integrals:
\[
\int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a)
\]
Area under a curve:
\[
\text{Area} = \int_a^b f(x) \, dx
\]
Worked Example 14.8 (Integration – non-calculator)
Integrate:
- \(\int 4x^3 \, dx\)
- \(\int (3x^2 + 2x - 5) \, dx\)
- \(\int \sin 3x \, dx\)
Solution:
(a) \(\int 4x^3 \, dx = x^4 + C\)
(b) \(\int (3x^2 + 2x - 5) \, dx = x^3 + x^2 - 5x + C\)
(c) \(\int \sin 3x \, dx = -\frac{1}{3}\cos 3x + C\)
Worked Example 14.9 (Definite integral – non-calculator)
Evaluate \(\int_0^2 (x^2 + 1) \, dx\).
Solution:
\(\int_0^2 (x^2 + 1) \, dx = \left[\frac{x^3}{3} + x\right]_0^2 = \left(\frac{8}{3} + 2\right) - 0 = \frac{8}{3} + \frac{6}{3} = \frac{14}{3}\)
Worked Example 14.10 (Area under curve – non-calculator)
Find the area under the curve \(y = x^2\) from \(x = 0\) to \(x = 3\).
Solution:
\(\text{Area} = \int_0^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9\)
14.6 Kinematics
Kinematics is the study of motion of a particle in a straight line.
Definitions:
- Displacement: \(s\) (position relative to a fixed point)
- Velocity: \(v = \frac{ds}{dt}\) (rate of change of displacement)
- Acceleration: \(a = \frac{dv}{dt} = \frac{d^2s}{dt^2}\) (rate of change of velocity)
\(v = \frac{ds}{dt}\)
\(a = \frac{dv}{dt} = \frac{d^2s}{dt^2}\)
Graphs:
- s-t graph: gradient = velocity
- v-t graph: gradient = acceleration; area under graph = displacement
Worked Example 14.11 (Mauritian context – kinematics)
The displacement of a particle moving in a straight line is given by \(s = t^3 - 6t^2 + 9t + 2\), where \(s\) is in metres and \(t\) is in seconds.
- Find the velocity and acceleration functions.
- Find the times when the particle is at rest.
- Find the displacement at these times.
- Find the total distance travelled in the first 3 seconds.
Solution:
(a) \(v = \frac{ds}{dt} = 3t^2 - 12t + 9\)
\(a = \frac{dv}{dt} = 6t - 12\)
(b) At rest: \(v = 0\)
\(3t^2 - 12t + 9 = 0 \Rightarrow 3(t^2 - 4t + 3) = 0 \Rightarrow 3(t - 1)(t - 3) = 0\)
\(t = 1\) or \(t = 3\) seconds
(c) \(s(1) = 1 - 6 + 9 + 2 = 6\) m
\(s(3) = 27 - 54 + 27 + 2 = 2\) m
(d) Total distance = \(|s(1) - s(0)| + |s(3) - s(1)|\)
\(s(0) = 2\)
\(|6 - 2| + |2 - 6| = 4 + 4 = 8\) m
Chapter 14 Summary
| Concept |
Key points |
| Differentiation |
\(\frac{dy}{dx} =\) rate of change; gradient of tangent |
| Standard derivatives |
\(x^n, \sin x, \cos x, \tan x, e^x, \ln x\) |
| Product rule |
\(\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\) |
| Quotient rule |
\(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\) |
| Chain rule |
\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\) |
| Stationary points |
\(f'(x) = 0\); use \(f''\) to classify |
| Connected rates |
\(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\) |
| Integration |
Reverse of differentiation |
| Power rule for integration |
\(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) |
| Definite integrals |
Area under curve: \(\int_a^b f(x) dx\) |
| Kinematics |
\(v = \frac{ds}{dt}\), \(a = \frac{dv}{dt}\) |
Exercises – Chapter 14
Easy (Drill – Non-Calculator)
- Differentiate:
(a) \(y = 3x^4\)
(b) \(y = x^5 + 2x^3 - 7x + 1\)
(c) \(y = \sin 2x\)
- Find the gradient of the curve \(y = x^3 - 4x\) at the point where \(x = 2\).
- Integrate:
(a) \(\int 4x^3 \, dx\)
(b) \(\int (3x^2 + 2x - 5) \, dx\)
(c) \(\int \sin 3x \, dx\)
- Find the area under the curve \(y = x^2\) from \(x = 0\) to \(x = 3\).
Medium (Examination Style)
- Differentiate:
(a) \(y = x^2 e^x\)
(b) \(y = \frac{x}{x^2 + 1}\)
(c) \(y = \ln(x^2 + 1)\)
- Find the equation of the tangent to the curve \(y = x^2 - 4x + 3\) at the point where \(x = 3\).
- Find the stationary points of \(y = x^3 - 6x^2 + 9x - 1\) and determine their nature.
- Mauritian context: The volume of a water tank is \(V = \frac{4}{3}\pi r^3\). Water is flowing into the tank at a rate of 2 m³/min. Find the rate at which the radius is increasing when \(r = 2\) m.
- Evaluate:
(a) \(\int_0^2 (x^2 + 1) \, dx\)
(b) \(\int_0^{\pi/2} \cos x \, dx\)
Hard (Challenge for A*)
- The curve \(y = f(x)\) passes through the point \((0, 1)\) and has gradient \(\frac{dy}{dx} = 3x^2 + 2x\). Find \(f(x)\).
- Mauritian context: The displacement of a particle moving in a straight line is given by \(s = t^3 - 6t^2 + 9t + 2\), where \(s\) is in metres and \(t\) is in seconds.
(a) Find the velocity and acceleration functions.
(b) Find the times when the particle is at rest.
(c) Find the displacement at these times.
(d) Find the total distance travelled in the first 3 seconds.
- Find the area enclosed between the curve \(y = x^2 - 4x + 3\) and the x-axis.
- A rectangular box with a square base has a fixed surface area of 150 cm². Find the maximum volume of the box.