Differentiation
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective
Understand the derivative from first principles, master the power rule for differentiating polynomial functions, and apply differentiation to determine gradients, tangents, and turning points.
Kenyan Context: Imagine a matatu accelerating away from the Machakos Country Bus terminal onto Mombasa Road. Its displacement from the stage after \(t\) seconds is recorded along a continuous curve. An average speed calculated over 10 minutes does not tell you whether the driver is exceeding the 80 km/h speed limit at a speed-trap camera at one exact instant. Differentiation is the mathematical tool that provides that instantaneous velocity \(\frac{ds}{dt}\) right at that split second.
Geometric Meaning: Consider a secant line connecting two points \(P(x, f(x))\) and \(Q(x+h, f(x+h))\) on a curve. As the step \(h\) decreases toward zero (\(h \to 0\)), the point \(Q\) slides down the curve toward \(P\). The secant line gradually rotates until it becomes the tangent line that touches the curve at exactly point \(P\). The gradient of this tangent line represents the instantaneous derivative \(f'(x)\).
The Power Rule: By applying the definition \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\) to \(f(x) = x^n\), we obtain the fundamental power rule: bring the exponent down to multiply the coefficient, then reduce the power by 1: \[\frac{d}{dx}\left(a x^n\right) = a \cdot n x^{n-1}\]
Use the interactive widget below to vary \(x_0\) and observe what happens as interval \(h\) shrinks towards 0. Watch the red secant line merge onto the green tangent line!
Key Formulas
The limit of the secant gradient as the interval \(h\) approaches zero gives the exact derivative.
Multiply by the existing power \(n\), then subtract \(1\) from the power.
A constant has no rate of change (gradient = 0). A linear term has a constant gradient \(a\).
Differentiate each term independently.
Points where the tangent line is perfectly horizontal.
- Local Maximum: \(\frac{d^2y}{dx^2} < 0\) (gradient changes from positive to negative).
- Local Minimum: \(\frac{d^2y}{dx^2} > 0\) (gradient changes from negative to positive).
Velocity \(v\) is the rate of change of displacement \(s\); acceleration \(a\) is the rate of change of velocity.
Worked Examples
Example 1: Gradient of a Quadratic Curve (Easy)
Find the gradient of the curve \(y = 3x^2 - 5x + 2\) at the point where \(x = 4\).
- Differentiate term-by-term:
\[\frac{dy}{dx} = 3(2x^{2-1}) - 5(1x^{1-1}) + 0 = 6x - 5\] - Substitute \(x = 4\) into the derivative expression:
\[\text{Gradient } m = 6(4) - 5 = 24 - 5 = 19\]
Answer: \(19\)
Example 2: Stationary Points and Curve Characteristics (Medium)
Find the coordinates of the turning points on the cubic curve \(y = 2x^3 - 3x^2 - 12x + 5\), and determine their nature.
- Differentiate the function:
\[\frac{dy}{dx} = 6x^2 - 6x - 12\] - Set \(\frac{dy}{dx} = 0\) for stationary points:
\[6x^2 - 6x - 12 = 0 \implies x^2 - x - 2 = 0\]
\[(x - 2)(x + 1) = 0 \implies x = 2 \quad \text{or} \quad x = -1\] - Find corresponding \(y\)-values:
When \(x = 2\): \(y = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15\). Point: \((2, -15)\).
When \(x = -1\): \(y = 2(-1) - 3(1) - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12\). Point: \((-1, 12)\). - Determine nature using the second derivative:
\[\frac{d^2y}{dx^2} = 12x - 6\]
At \(x = -1\): \(\frac{d^2y}{dx^2} = 12(-1) - 6 = -18 < 0 \implies\) Local Maximum at \((-1, 12)\).
At \(x = 2\): \(\frac{d^2y}{dx^2} = 12(2) - 6 = 18 > 0 \implies\) Local Minimum at \((2, -15)\).
Example 3: Optimization / Real-World Modeling (Hard)
A tea farmer in Kericho has \(120\text{ m}\) of wire mesh to fence off a rectangular nursery divided into two equal compartments by an interior fence parallel to the width \(w\). Find the maximum possible total area of the nursery in square metres.
- Formulate the constraint equation:
The perimeter with the internal divider requires \(2\) lengths \(L\) and \(3\) widths \(w\):
\[2L + 3w = 120 \implies 2L = 120 - 3w \implies L = 60 - 1.5w\] - Express total area \(A\) as a single-variable function:
\[A = L \cdot w = (60 - 1.5w)w = 60w - 1.5w^2\] - Differentiate and find stationary point:
\[\frac{dA}{dw} = 60 - 3w\]
For maximum area, set \(\frac{dA}{dw} = 0\):
\[60 - 3w = 0 \implies 3w = 60 \implies w = 20\text{ m}\] - Check second derivative:
\[\frac{d^2A}{dw^2} = -3 < 0 \quad (\text{Confirms maximum})\] - Compute the maximum area:
\[A_{\max} = 60(20) - 1.5(20^2) = 1200 - 1.5(400) = 1200 - 600 = 600\text{ m}^2\]
Answer: \(600\text{ m}^2\)
Common Mistakes
Misconception 1: Differentiating a constant into the same constant.
Incorrect: Differentiating \(y = 5x^3 + 8\) to get \(\frac{dy}{dx} = 15x^2 + 8\).
Correction: The derivative of any constant \(c\) is \(0\). Thus, \(\frac{dy}{dx} = 15x^2\).
Incorrect: Writing \(\frac{d}{dx}(x^4) = 4x^4\) or \(\frac{d}{dx}(x) = 1x = x\).
Correction: You must always reduce the power by 1: \(\frac{d}{dx}(x^4) = 4x^3\) and \(\frac{d}{dx}(x^1) = 1x^0 = 1\).
Incorrect: Solving \(y = 0\) to locate turning points.
Correction: \(y = 0\) gives where the graph crosses the \(x\)-axis (roots/intercepts). Turning points occur where the tangent is horizontal, which is found strictly by solving \(\frac{dy}{dx} = 0\).
Real World
Practice