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Graphs & Basic Differentiation

Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.

N/A Pathway: N/A

First Principles

Objective: Sketch function graphs and apply basic differentiation rules to calculate gradients and turning points.

On a curved graph, the gradient continuously changes from point to point. Differentiation provides the exact instantaneous rate of change (gradient) at any point \( x \) by finding the slope of the tangent line touching the curve.

(a) The Derivative as Tangent Gradient

For a function \( y = f(x) \), the derivative \( \frac{dy}{dx} \) (or \( f'(x) \)) measures the rate of change of \( y \) with respect to \( x \).

(b) The Power Rule

For any term \( a x^n \), its derivative is found by multiplying by the exponent \( n \) and reducing the power by 1:

\[\frac{d}{dx}\left(a x^n\right) = a \cdot n x^{n-1}\]

Note: Constant terms differentiate to zero (\( \frac{d}{dx}(c) = 0 \)).

Turning Point (dy/dx = 0) Positive Gradient (dy/dx > 0)

Key Formulas

\[\frac{d}{dx}\left(x^n\right) = n x^{n-1}\] — Power Rule: Multiply by exponent \( n \) and subtract 1 from the power.
\[\frac{d}{dx}\left(a x^n + b x + c\right) = a n x^{n-1} + b\] — Polynomial Differentiation: Differentiate term by term; constant \( c \) vanishes.
\[m = \left.\frac{dy}{dx}\right|_{x = x_0}\] — Gradient at a Point: Substitute \( x_0 \) into derivative \( \frac{dy}{dx} \).
\[\frac{dy}{dx} = 0\] — Stationary / Turning Point Condition: Tangent is horizontal at local maximums and minimums.

Worked Examples

Example 1 (Easy - Straight Line Gradient): A line passes through \( (4, 10) \) with a y-intercept of \( 2 \). Calculate the gradient \( m \).
  1. Use equation form \( y = mx + c \) with \( c = 2 \).
  2. Substitute \( (4, 10) \): \( 10 = 4m + 2 \).
  3. Solve for \( m \): \( 4m = 8 \implies m = 2 \).

Answer: 2

Example 2 (Medium - Derivative at a Specific Point): Determine the gradient of the curve \( y = 2x^2 - 3x + 5 \) at \( x = 2 \).
  1. Differentiate term by term: \( \frac{dy}{dx} = \frac{d}{dx}(2x^2) - \frac{d}{dx}(3x) + \frac{d}{dx}(5) = 4x - 3 \).
  2. Substitute \( x = 2 \): \( \frac{dy}{dx} = 4(2) - 3 = 8 - 3 = 5 \).

Answer: 5

Example 3 (Hard - Maximizing Profit Function): A local vendor in Mombasa models his daily profit (in KSh thousands) from selling nyama choma by \( P(q) = -q^2 + 8q - 5 \), where \( q \) is the number of goats roasted. Calculate the output \( q \) that maximizes profit.
  1. Differentiate \( P(q) \): \( P'(q) = -2q + 8 \).
  2. Set \( P'(q) = 0 \) for maximum turning point: \( -2q + 8 = 0 \implies 2q = 8 \implies q = 4 \).
  3. Verify second derivative: \( P''(q) = -2 < 0 \) (concave down \( \implies \) maximum).

Answer: 4

Common Mistakes

Mistake Confusing the y-intercept \( c \) with the gradient \( m \) in \( y = mx + c \).
Correction The gradient \( m \) measures steepness (rise/run), while \( c \) is the height where the graph crosses the vertical axis.
Why it feels right Both are fixed numerical constants in linear equations.
Mistake Differentiating constant terms as if they contain \( x \) (e.g. \( \frac{d}{dx}(7) = 7 \)).
Correction Constants have zero rate of change, so \( \frac{d}{dx}(c) = 0 \).
Why it feels right Students expect every number in an expression to carry over into the derivative.

Real World

Marginal Cost & Revenue in Business: Market stall owners in Gikomba use derivatives to compute marginal profit \( — \) the exact profit added by selling one additional item.
Vehicle Kinematics: Traffic engineers differentiate displacement functions \( s(t) \) of matatus on Thika Superhighway to model instantaneous velocity \( v(t) \).

Practice

The straight line y = mx + c passes through the point (4, 10) and has a y-intercept of 2. What is the value of the gradient m? (Type only the number, e.g., 2)
Review the concepts above.
What is the gradient of the straight line that passes through the points (2, 5) and (8, -7)? (Type only the number, e.g., -2)
Review the concepts above.
For the curve y = x^3 - 5x + 7, calculate the gradient dy/dx at x = 2. (Type only the number, e.g., 7)
Review the concepts above.
Determine the gradient of the curve f(x) = 2x^2 - 3x + 5 at the point where x = 2. (Type only the number, e.g., 5)
Review the concepts above.
A profit function is given by P(q) = -q^2 + 40q - 300. Calculate the value of q that maximizes profit (where dP/dq = 0). (Type only the number, e.g., 20)
Review the concepts above.
A parabola has equation y = x^2 - 6x + 8. At what value of x is the gradient equal to 4? (Type only the number, e.g., 5)
Review the concepts above.